Differentiation of implicit function Questions in English

(D) Given equation: $2x \sqrt{y} + 2y \sqrt{x} = 4x \sqrt{x} + y \sqrt{y}$.
Divide the entire equation by $x \sqrt{x}$ (assuming $x > 0$):
$2 \frac{\sqrt{y}}{\sqrt{x}} + 2 \frac{y}{x} = 4 + \frac{y \sqrt{y}}{x \sqrt{x}}$.
Let $v = \sqrt{\frac{y}{x}}$,then $v^2 = \frac{y}{x}$.
The equation becomes $2v + 2v^2 = 4 + v^3$,which is $v^3 - 2v^2 - 2v + 4 = 0$.
Factoring: $v^2(v - 2) - 2(v - 2) = 0 \Rightarrow (v^2 - 2)(v - 2) = 0$.
At $(1, 4)$,$v = \sqrt{\frac{4}{1}} = 2$,which satisfies the equation.
For a homogeneous equation of the form $f(\frac{y}{x}) = c$,the derivative is $\frac{dy}{dx} = \frac{y}{x}$.
At $(1, 4)$,the slope is $\frac{dy}{dx} = \frac{4}{1} = 4$.

(A) Given the equation: $xe^{xy} = y + e^{\sin 2x}$.
First,find the value of $y$ when $x = 0$:
Substituting $x = 0$ into the equation: $0 \cdot e^0 = y + e^{\sin 0} \implies 0 = y + e^0 \implies 0 = y + 1 \implies y = -1$.
Differentiating both sides with respect to $x$ using the product rule and chain rule:
$\frac{d}{dx}(xe^{xy}) = \frac{d}{dx}(y + e^{\sin 2x})$
$e^{xy} + x \cdot e^{xy} \cdot (y + x \frac{dy}{dx}) = \frac{dy}{dx} + e^{\sin 2x} \cdot \cos 2x \cdot 2$.
Now,substitute $x = 0$ and $y = -1$ into the differentiated equation:
$e^{0(-1)} + 0 \cdot e^{0(-1)} \cdot (-1 + 0 \cdot \frac{dy}{dx}) = \frac{dy}{dx} + e^{\sin 0} \cdot \cos 0 \cdot 2$
$1 + 0 = \frac{dy}{dx} + 1 \cdot 1 \cdot 2$
$1 = \frac{dy}{dx} + 2$
$\frac{dy}{dx} = 1 - 2 = -1$.

(A) Given the equation $f(x) \cdot f(f(x)) = x^2 + 1$.
Substitute $x = 1$ into the equation:
$f(1) \cdot f(f(1)) = 1^2 + 1$
Since $f(1) = 2$,we have $2 \cdot f(2) = 2$,which implies $f(2) = 1$.
Differentiate both sides of the original equation with respect to $x$ using the product rule and chain rule:
$f'(x) \cdot f(f(x)) + f(x) \cdot f'(f(x)) \cdot f'(x) = 2x$.
Substitute $x = 1$ into the derivative equation:
$f'(1) \cdot f(f(1)) + f(1) \cdot f'(f(1)) \cdot f'(1) = 2(1)$.
Using $f(1) = 2$,$f'(1) = k$,and $f(2) = 1$:
$k \cdot f(2) + 2 \cdot f'(2) \cdot k = 2$.
Substitute $f(2) = 1$:
$k(1) + 2k \cdot f'(2) = 2$.
Solve for $f'(2)$:
$2k \cdot f'(2) = 2 - k$
$f'(2) = \frac{2 - k}{2k} = \frac{2}{2k} - \frac{k}{2k} = \frac{1}{k} - \frac{1}{2}$.

(B) Given the equation: $\sin(x + y) + \cos(2x + 2y) = \ln(3(x + y))$.
Differentiating both sides with respect to $x$ using the chain rule:
$\cos(x + y) \cdot (1 + \frac{dy}{dx}) - \sin(2x + 2y) \cdot (2 + 2\frac{dy}{dx}) = \frac{1}{3(x + y)} \cdot 3(1 + \frac{dy}{dx})$.
Simplifying the right side:
$\cos(x + y) \cdot (1 + \frac{dy}{dx}) - 2\sin(2x + 2y) \cdot (1 + \frac{dy}{dx}) = \frac{1}{x + y} \cdot (1 + \frac{dy}{dx})$.
Rearranging the terms:
$(1 + \frac{dy}{dx}) [\cos(x + y) - 2\sin(2x + 2y) - \frac{1}{x + y}] = 0$.
Since the term in the square bracket is not identically zero,we must have:
$1 + \frac{dy}{dx} = 0$.
Therefore,$\frac{dy}{dx} = -1$.

(B) Given the curve equation: $y^3 - 3xy + 2 = 0$.
Differentiating with respect to $x$:
$3y^2 \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} = 0$
$\frac{dy}{dx}(3y^2 - 3x) = 3y$
$\frac{dy}{dx} = \frac{y}{y^2 - x}$.
For the tangent to be horizontal,$\frac{dy}{dx} = 0$,which implies $y = 0$.
Substituting $y = 0$ into the curve equation: $0^3 - 3x(0) + 2 = 0 \Rightarrow 2 = 0$,which is impossible.
Thus,there are no points where the tangent is horizontal,so $H = \phi$.
For the tangent to be vertical,$\frac{dy}{dx} = \infty$,which implies the denominator $y^2 - x = 0$,so $x = y^2$.
Substituting $x = y^2$ into the curve equation: $y^3 - 3(y^2)y + 2 = 0$
$y^3 - 3y^3 + 2 = 0$
$-2y^3 + 2 = 0 \Rightarrow y^3 = 1 \Rightarrow y = 1$.
Since $x = y^2$,$x = 1^2 = 1$.
Thus,$V = \{(1, 1)\}$.

(C) Given $(x - y) f(x + y) - (x + y) f(x - y) = 4xy(x^2 - y^2)$.
Dividing by $(x^2 - y^2) = (x+y)(x-y)$,we get $\frac{f(x+y)}{x+y} - \frac{f(x-y)}{x-y} = 4xy$.
Let $u = x+y$ and $v = x-y$. Then $x = \frac{u+v}{2}$ and $y = \frac{u-v}{2}$.
So $4xy = 4(\frac{u+v}{2})(\frac{u-v}{2}) = u^2 - v^2$.
Thus,$\frac{f(u)}{u} - \frac{f(v)}{v} = u^2 - v^2 \Rightarrow \frac{f(u)}{u} - u^2 = \frac{f(v)}{v} - v^2 = c$.
Therefore,$f(u) = cu + u^3$. Given $f(1) = 2$,we have $c(1) + 1^3 = 2 \Rightarrow c = 1$.
So $f(x) = x + x^3$.
The inequality becomes $\frac{|x^3|^{1/3}}{17} + \frac{|y^3|^{1/3}}{2} \le \frac{1}{4} \Rightarrow \frac{|x|}{17} + \frac{|y|}{2} \le \frac{1}{4}$.
This represents a rhombus with vertices $(\pm \frac{17}{4}, 0)$ and $(0, \pm \frac{1}{2})$.
Area $= 4 \times \text{Area of triangle in first quadrant} = 4 \times (\frac{1}{2} \times \frac{17}{4} \times \frac{1}{2}) = \frac{17}{4}$.
Since $f(4) = 4 + 4^3 = 68$,we have $\frac{f(4)}{16} = \frac{68}{16} = \frac{17}{4}$.
Thus,the area is $\frac{f(4)}{16}$.

(B) Given the relation: $4x{e^{xy}} = y + 5{\sin ^2}x$.
First,find the value of $y$ at $x=0$.
Substituting $x=0$ into the equation: $4(0){e^{0}} = y + 5{\sin ^2}(0) \implies 0 = y + 0 \implies y(0) = 0$.
Now,differentiate both sides with respect to $x$ using the product rule and chain rule:
$\frac{d}{dx}(4x{e^{xy}}) = \frac{d}{dx}(y + 5{\sin ^2}x)$
$4{e^{xy}} + 4x{e^{xy}}(y + x y') = y' + 10\sin x \cos x$.
Substitute $x=0$ and $y=0$ into the differentiated equation:
$4{e^{0}} + 4(0){e^{0}}(0 + 0 \cdot y'(0)) = y'(0) + 10\sin(0)\cos(0)$.
$4(1) + 0 = y'(0) + 0$.
Therefore,$y'(0) = 4$.

(C) Given the equation: $\ln \left( {(e - 1){e^{xy}} + {x^2}} \right) = {x^2} + {y^2}$.
First,we differentiate both sides with respect to $x$ using the chain rule:
$\frac{1}{(e - 1){e^{xy}} + {x^2}} \cdot \left( {(e - 1){e^{xy}} \cdot \left( y + x \frac{dy}{dx} \right) + 2x} \right) = 2x + 2y \frac{dy}{dx}$.
Now,substitute the point $(x, y) = (1, 0)$ into the equation:
At $x = 1, y = 0$,the term inside the logarithm is $(e - 1){e^0} + 1^2 = e - 1 + 1 = e$.
So,$\ln(e) = 1^2 + 0^2 = 1$,which is consistent.
Substituting $(1, 0)$ into the differentiated equation:
$\frac{1}{e} \cdot \left( {(e - 1){e^0} \cdot \left( 0 + 1 \cdot \frac{dy}{dx} \right) + 2(1)} \right) = 2(1) + 2(0) \cdot \frac{dy}{dx}$.
$\frac{1}{e} \cdot \left( {(e - 1) \frac{dy}{dx} + 2} \right) = 2$.
$(e - 1) \frac{dy}{dx} + 2 = 2e$.
$(e - 1) \frac{dy}{dx} = 2e - 2$.
$(e - 1) \frac{dy}{dx} = 2(e - 1)$.
Therefore,$\left. \frac{dy}{dx} \right|_{(1,0)} = 2$.

(D) Given the functional equation: $f(x + 2y) = 2yf(x) + xf(y) - 3xy + 1$.
Setting $x = 0$ and $y = 0$,we get $f(0) = 0 + 0 - 0 + 1$,so $f(0) = 1$.
To find $f'(x)$,we differentiate the given equation with respect to $x$ keeping $y$ constant:
$f'(x + 2y) = 2yf'(x) + f(y) - 3y$.
Setting $x = 0$ in this derivative equation:
$f'(2y) = 2yf'(0) + f(y) - 3y$.
Since $f'(0) = 1$,we have $f'(2y) = 2y(1) + f(y) - 3y = f(y) - y$.
Let $2y = t$,then $y = t/2$. Thus,$f'(t) = f(t/2) - t/2$.
Alternatively,differentiate the original equation with respect to $y$ keeping $x$ constant:
$2f'(x + 2y) = 2f(x) + xf'(y) - 3x$.
Setting $y = 0$:
$2f'(x) = 2f(x) + xf'(0) - 3x$.
Since $f'(0) = 1$,we have $2f'(x) = 2f(x) + x - 3x$,which simplifies to $2f'(x) = 2f(x) - 2x$,or $f'(x) - f(x) = -x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = -x$.
The integrating factor is $IF = e^{\int -1 dx} = e^{-x}$.
Multiplying by $IF$: $e^{-x} f'(x) - e^{-x} f(x) = -x e^{-x}$.
Integrating both sides: $\int \frac{d}{dx} (f(x) e^{-x}) dx = \int -x e^{-x} dx$.
$f(x) e^{-x} = -[x(-e^{-x}) - \int 1(-e^{-x}) dx] = x e^{-x} + e^{-x} + C$.
$f(x) = x + 1 + Ce^x$.
Using $f(0) = 1$: $1 = 0 + 1 + C(1) \Rightarrow C = 0$.
Thus,$f(x) = x + 1$.
Therefore,$f(2) = 2 + 1 = 3$.

(A) Given $f'(x) = f^2(x)$ and $f(0) = 0$.
If $f(x)$ is not identically zero,then $\frac{f'(x)}{f^2(x)} = 1$.
Integrating both sides,we get $-\frac{1}{f(x)} = x + c$,which implies $f(x) = -\frac{1}{x+c}$.
However,$f(0) = 0$ implies $-\frac{1}{c} = 0$,which is impossible.
Thus,the only twice differentiable function satisfying $f(0) = 0$ and $f'(x) = f^2(x)$ is $f(x) = 0$ for all $x \in R$.
If $f(x) = 0$,then $f'(x) = 0$ and $f''(x) = 0$.
Therefore,$\lim_{x \to 2} (f(x) + xf'(x) + x^2f''(x)) = 0 + 2(0) + 4(0) = 0$.
Checking the options for $f(x) = 0$:
$A: 2(0) + 4(0) = 0$.
$B: 2(0) + 4(0) - 0 = 0$.
$C: -1 + 0 + 0 = -1$.
$D: 1 + 0 + 0 = 1$.
Since $f(x) = 0$ is the solution,the expression evaluates to $0$. Option $A$ and $B$ both result in $0$. Given the structure,$A$ is the most simplified form.

(B) Given the equation $y = x^{x^{x...\infty}}$.
Since the exponent is infinite,we can write $y = x^y$.
Taking the natural logarithm on both sides,we get $\log y = \log(x^y) = y \log x$.
Differentiating both sides with respect to $x$,we get $\frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \log x + y \cdot \frac{1}{x}$.
Multiplying the entire equation by $xy$,we get $x \frac{dy}{dx} = xy \log x \frac{dy}{dx} + y^2$.
Rearranging the terms,we get $x \frac{dy}{dx} - xy \log x \frac{dy}{dx} = y^2$.
Factoring out $x \frac{dy}{dx}$,we get $x \frac{dy}{dx} (1 - y \log x) = y^2$.
Thus,$x (1 - y \log x) \frac{dy}{dx} = y^2$.

(A) Given the implicit equation $F(x, y) = {x^2}{e^y} + 2xy{e^x} + 13 = 0$.
Using the formula for the derivative of an implicit function,$\frac{dy}{dx} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$.
First,find the partial derivative with respect to $x$:
$\frac{\partial F}{\partial x} = 2x{e^y} + 2y{e^x} + 2xy{e^x}$.
Next,find the partial derivative with respect to $y$:
$\frac{\partial F}{\partial y} = {x^2}{e^y} + 2x{e^x}$.
Now,substitute these into the formula:
$\frac{dy}{dx} = -\frac{2x{e^y} + 2y{e^x} + 2xy{e^x}}{{x^2}{e^y} + 2x{e^x}}$.
Divide the numerator and denominator by ${e^x}$:
$\frac{dy}{dx} = -\frac{2x{e^{y-x}} + 2y + 2xy}{{x^2}{e^{y-x}} + 2x}$.
Factor out $2$ in the numerator and $x$ in the denominator:
$\frac{dy}{dx} = -\frac{2(x{e^{y-x}} + y(1+x))}{x(x{e^{y-x}} + 2)}$.
This matches option $A$.

(A) Given the infinite series $y = x^2 + \frac{1}{x^2 + \frac{1}{x^2 + \dots \infty}}$.
Since the series is infinite,we can observe that the expression repeats itself,so we can write: $y = x^2 + \frac{1}{y}$.
Multiplying both sides by $y$,we get: $y^2 = x^2 y + 1$.
Differentiating both sides with respect to $x$ using the product rule and chain rule:
$2y \frac{dy}{dx} = (x^2 \frac{dy}{dx} + y \cdot 2x) + 0$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$2y \frac{dy}{dx} - x^2 \frac{dy}{dx} = 2xy$.
$\frac{dy}{dx} (2y - x^2) = 2xy$.
Therefore,$\frac{dy}{dx} = \frac{2xy}{2y - x^2}$.

(A) Given the inequality $f'(x) \cos x \le f(x) \sin x$ for $x \ge 0$.
Rearranging the terms,we get $f'(x) \cos x - f(x) \sin x \le 0$.
This expression is the derivative of the product $f(x) \cos x$ with respect to $x$,i.e.,$\frac{d}{dx}(f(x) \cos x) \le 0$.
Let $g(x) = f(x) \cos x$. Since $g'(x) \le 0$,the function $g(x)$ is monotonically non-increasing for $x \ge 0$.
For $x \in [0, \pi/2]$,$\cos x \ge 0$. As $x$ approaches $\pi/2$ from the left,$g(x) = f(x) \cos x$ must remain non-negative because $f(x) \ge 0$ and $\cos x \ge 0$.
However,at $x = \pi/2$,$g(\pi/2) = f(\pi/2) \cdot 0 = 0$.
Since $g(x)$ is non-increasing and $g(x) \ge 0$ for $x \in [0, \pi/2]$,and $g(\pi/2) = 0$,it implies $g(x) = 0$ for all $x \ge \pi/2$.
Therefore,$f(2\pi) \cos(2\pi) = 0$,which means $f(2\pi) \cdot 1 = 0$,so $f(2\pi) = 0$.

(B) Given the equation: $x \ln(\ln x) - x^2 + y^2 = 4$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}[x \ln(\ln x)] - \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(4)$
Using the product rule on the first term: $1 \cdot \ln(\ln x) + x \cdot \frac{1}{\ln x} \cdot \frac{1}{x} - 2x + 2y \frac{dy}{dx} = 0$
$\ln(\ln x) + \frac{1}{\ln x} - 2x + 2y \frac{dy}{dx} = 0$
At $x = e$,$\ln(\ln e) = \ln(1) = 0$ and $\ln e = 1$:
$0 + \frac{1}{1} - 2e + 2y \frac{dy}{dx} = 0$
$1 - 2e + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{2e - 1}{2y}$
Now,find $y$ at $x = e$ from the original equation:
$e \ln(\ln e) - e^2 + y^2 = 4$
$e(0) - e^2 + y^2 = 4 \implies y^2 = 4 + e^2 \implies y = \sqrt{4 + e^2}$ (since $y > 0$)
Substituting $y$ into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{2e - 1}{2\sqrt{4 + e^2}}$

(C) Given the equation ${e^y} + xy = e$.
At $x = 0$,${e^y} + 0 = e \implies {e^y} = e \implies y = 1$.
Differentiating both sides with respect to $x$:
${e^y} \frac{dy}{dx} + x \frac{dy}{dx} + y = 0$.
At $(0, 1)$,${e^1} \frac{dy}{dx} + 0 + 1 = 0 \implies e \frac{dy}{dx} = -1 \implies \frac{dy}{dx} = -\frac{1}{e}$.
Now,differentiate ${e^y} \frac{dy}{dx} + x \frac{dy}{dx} + y = 0$ again with respect to $x$:
${e^y} \frac{d^2y}{dx^2} + {e^y} \left( \frac{dy}{dx} \right)^2 + x \frac{d^2y}{dx^2} + \frac{dy}{dx} + \frac{dy}{dx} = 0$.
Substitute $x = 0, y = 1, \frac{dy}{dx} = -\frac{1}{e}$:
$e \frac{d^2y}{dx^2} + e \left( -\frac{1}{e} \right)^2 + 0 + 2 \left( -\frac{1}{e} \right) = 0$.
$e \frac{d^2y}{dx^2} + \frac{1}{e} - \frac{2}{e} = 0$.
$e \frac{d^2y}{dx^2} = \frac{1}{e} \implies \frac{d^2y}{dx^2} = \frac{1}{e^2}$.
Thus,the ordered pair is $\left( -\frac{1}{e}, \frac{1}{e^2} \right)$.

(B) Let $x = \sin \theta$ and $y = \sin \alpha$.
Substituting these into the given equation $y \sqrt{1-x^{2}} = k - x \sqrt{1-y^{2}}$,we get:
$\sin \alpha \cos \theta = k - \sin \theta \cos \alpha$
$\sin \alpha \cos \theta + \cos \alpha \sin \theta = k$
$\sin(\alpha + \theta) = k$
$\alpha + \theta = \sin^{-1} k$
Substituting back,we have $\sin^{-1} y + \sin^{-1} x = \sin^{-1} k$.
Differentiating both sides with respect to $x$:
$\frac{1}{\sqrt{1-x^{2}}} + \frac{1}{\sqrt{1-y^{2}}} \frac{dy}{dx} = 0$
At $x = \frac{1}{2}$,$y = -\frac{1}{4}$.
$\sqrt{1-x^{2}} = \sqrt{1 - (\frac{1}{2})^{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
$\sqrt{1-y^{2}} = \sqrt{1 - (-\frac{1}{4})^{2}} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$.
Substituting these values:
$\frac{1}{\sqrt{3}/2} + \frac{1}{\sqrt{15}/4} \frac{dy}{dx} = 0$
$\frac{2}{\sqrt{3}} + \frac{4}{\sqrt{15}} \frac{dy}{dx} = 0$
$\frac{4}{\sqrt{15}} \frac{dy}{dx} = -\frac{2}{\sqrt{3}}$
$\frac{dy}{dx} = -\frac{2}{\sqrt{3}} \times \frac{\sqrt{15}}{4} = -\frac{1}{2} \times \sqrt{5} = -\frac{\sqrt{5}}{2}$.

(C) Given the equation $x^{k}+y^{k}=a^{k}$.
Differentiating both sides with respect to $x$,we get:
$k x^{k-1} + k y^{k-1} \frac{dy}{dx} = 0$.
Dividing by $k$ (since $k > 0$):
$x^{k-1} + y^{k-1} \frac{dy}{dx} = 0$.
Rearranging to solve for $\frac{dy}{dx}$:
$\frac{dy}{dx} = -\frac{x^{k-1}}{y^{k-1}} = -\left(\frac{x}{y}\right)^{k-1}$.
We are given $\frac{dy}{dx} + \left(\frac{y}{x}\right)^{\frac{1}{3}} = 0$,which implies $\frac{dy}{dx} = -\left(\frac{y}{x}\right)^{\frac{1}{3}} = -\left(\frac{x}{y}\right)^{-\frac{1}{3}}$.
Comparing the two expressions for $\frac{dy}{dx}$:
$-\left(\frac{x}{y}\right)^{k-1} = -\left(\frac{x}{y}\right)^{-\frac{1}{3}}$.
Equating the exponents:
$k - 1 = -\frac{1}{3}$.
Solving for $k$:
$k = 1 - \frac{1}{3} = \frac{2}{3}$.

(A) We differentiate the given equation $y+\sin y=\cos x$ with respect to $x$:
$\frac{d}{dx}(y) + \frac{d}{dx}(\sin y) = \frac{d}{dx}(\cos x)$
Using the chain rule for $\frac{d}{dx}(\sin y)$:
$\frac{dy}{dx} + \cos y \cdot \frac{dy}{dx} = -\sin x$
Factor out $\frac{dy}{dx}$:
$\frac{dy}{dx}(1 + \cos y) = -\sin x$
Therefore,the derivative is:
$\frac{dy}{dx} = -\frac{\sin x}{1+\cos y}$
where $y \neq (2n+1)\pi$.

(A) Given the equation: $2x + 3y = \sin x$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(2x + 3y) = \frac{d}{dx}(\sin x)$
Applying the sum rule and derivative rules:
$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \cos x$
$2 + 3 \frac{dy}{dx} = \cos x$
Subtracting $2$ from both sides:
$3 \frac{dy}{dx} = \cos x - 2$
Dividing by $3$:
$\frac{dy}{dx} = \frac{\cos x - 2}{3}$

(A) The given equation is $2x + 3y = \sin y$.
Differentiating both sides with respect to $x$,we get:
$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y)$
$2 + 3 \frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$2 = \cos y \cdot \frac{dy}{dx} - 3 \frac{dy}{dx}$
$2 = (\cos y - 3) \frac{dy}{dx}$
Therefore,$\frac{dy}{dx} = \frac{2}{\cos y - 3}$.

(A) The given equation is $ax + by^2 = \cos y$.
Differentiating both sides with respect to $x$,we get:
$\frac{d}{dx}(ax) + \frac{d}{dx}(by^2) = \frac{d}{dx}(\cos y)$
Using the chain rule,we have:
$a + b(2y \frac{dy}{dx}) = -\sin y \frac{dy}{dx}$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$a = -\sin y \frac{dy}{dx} - 2by \frac{dy}{dx}$
$a = -\frac{dy}{dx}(2by + \sin y)$
Therefore,$\frac{dy}{dx} = \frac{-a}{2by + \sin y}$.

(A) The given equation is $xy + y^2 = \tan x + y$.
Differentiating both sides with respect to $x$,we get:
$\frac{d}{dx}(xy + y^2) = \frac{d}{dx}(\tan x + y)$
Using the product rule for $xy$ and the chain rule for $y^2$:
$y \cdot \frac{d}{dx}(x) + x \cdot \frac{dy}{dx} + 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$
$y + x \frac{dy}{dx} + 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$
Grouping the terms containing $\frac{dy}{dx}$ on one side:
$x \frac{dy}{dx} + 2y \frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x - y$
$(x + 2y - 1) \frac{dy}{dx} = \sec^2 x - y$
Therefore,$\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}$.

(A) The given equation is $x^{2}+xy+y^{2}=100$.
Differentiating both sides with respect to $x$,we get:
$\frac{d}{dx}(x^{2}+xy+y^{2}) = \frac{d}{dx}(100)$
Using the sum rule and the chain rule:
$\frac{d}{dx}(x^{2}) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^{2}) = 0$
Applying the product rule to $xy$:
$2x + (y \cdot 1 + x \cdot \frac{dy}{dx}) + 2y \frac{dy}{dx} = 0$
Grouping the $\frac{dy}{dx}$ terms:
$2x + y + (x + 2y) \frac{dy}{dx} = 0$
Solving for $\frac{dy}{dx}$:
$(x + 2y) \frac{dy}{dx} = -(2x + y)$
$\frac{dy}{dx} = -\frac{2x+y}{x+2y}$

(A) The given equation is $x^{3}+x^{2}y+xy^{2}+y^{3}=81$.
Differentiating both sides with respect to $x$,we get:
$\frac{d}{dx}(x^{3}+x^{2}y+xy^{2}+y^{3}) = \frac{d}{dx}(81)$
Using the sum rule and product rule:
$\frac{d}{dx}(x^{3}) + \frac{d}{dx}(x^{2}y) + \frac{d}{dx}(xy^{2}) + \frac{d}{dx}(y^{3}) = 0$
$3x^{2} + (x^{2}\frac{dy}{dx} + y(2x)) + (x(2y\frac{dy}{dx}) + y^{2}(1)) + 3y^{2}\frac{dy}{dx} = 0$
Grouping the terms containing $\frac{dy}{dx}$:
$(x^{2} + 2xy + 3y^{2})\frac{dy}{dx} + (3x^{2} + 2xy + y^{2}) = 0$
Solving for $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{-(3x^{2} + 2xy + y^{2})}{x^{2} + 2xy + 3y^{2}}$

(A) The given relationship is $\sin^{2} y + \cos(xy) = \pi$.
Differentiating both sides with respect to $y$,we get:
$\frac{d}{dy}(\sin^{2} y) + \frac{d}{dy}(\cos(xy)) = \frac{d}{dy}(\pi)$
Using the chain rule:
$\frac{d}{dy}(\sin^{2} y) = 2 \sin y \cos y \frac{dy}{dy} = \sin(2y)$.
For the second term:
$\frac{d}{dy}(\cos(xy)) = -\sin(xy) \cdot \frac{d}{dy}(xy) = -\sin(xy) \cdot (x + y \frac{dx}{dy})$.
Substituting these into the equation:
$\sin(2y) - \sin(xy) \cdot (x + y \frac{dx}{dy}) = 0$.
Rearranging to solve for $\frac{dx}{dy}$:
$\sin(2y) - x \sin(xy) - y \sin(xy) \frac{dx}{dy} = 0$.
$y \sin(xy) \frac{dx}{dy} = \sin(2y) - x \sin(xy)$.
Therefore,$\frac{dx}{dy} = \frac{\sin(2y) - x \sin(xy)}{y \sin(xy)}$.

(A) Given the equation: $\sin^{2} x + \cos^{2} y = 1$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\sin^{2} x) + \frac{d}{dx}(\cos^{2} y) = \frac{d}{dx}(1)$
Using the chain rule:
$2 \sin x \cos x + 2 \cos y (-\sin y) \frac{dy}{dx} = 0$
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$:
$\sin 2x - \sin 2y \frac{dy}{dx} = 0$
Rearranging to solve for $\frac{dy}{dx}$:
$\sin 2y \frac{dy}{dx} = \sin 2x$
$\frac{dy}{dx} = \frac{\sin 2x}{\sin 2y}$

(A) Given that $y^{x}+x^{y}+x^{x}=a^{b}$.
Let $u=y^{x}, v=x^{y}$ and $w=x^{x}$,then $u+v+w=a^{b}$.
Differentiating with respect to $x$,we get $\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}=0$ ... $(1)$.
For $u=y^{x}$,taking log on both sides: $\log u = x \log y$.
Differentiating: $\frac{1}{u} \frac{du}{dx} = x \cdot \frac{1}{y} \frac{dy}{dx} + \log y \cdot 1 \implies \frac{du}{dx} = y^{x} \left( \frac{x}{y} \frac{dy}{dx} + \log y \right)$ ... $(2)$.
For $v=x^{y}$,taking log on both sides: $\log v = y \log x$.
Differentiating: $\frac{1}{v} \frac{dv}{dx} = y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx} \implies \frac{dv}{dx} = x^{y} \left( \frac{y}{x} + \log x \frac{dy}{dx} \right)$ ... $(3)$.
For $w=x^{x}$,taking log on both sides: $\log w = x \log x$.
Differentiating: $\frac{1}{w} \frac{dw}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1 \implies \frac{dw}{dx} = x^{x}(1 + \log x)$ ... $(4)$.
Substituting $(2), (3), (4)$ into $(1)$:
$y^{x} \left( \frac{x}{y} \frac{dy}{dx} + \log y \right) + x^{y} \left( \frac{y}{x} + \log x \frac{dy}{dx} \right) + x^{x}(1 + \log x) = 0$.
Grouping $\frac{dy}{dx}$ terms: $\frac{dy}{dx} (x \cdot y^{x-1} + x^{y} \log x) = -[y^{x} \log y + y \cdot x^{y-1} + x^{x}(1 + \log x)]$.
Thus,$\frac{dy}{dx} = \frac{-[y^{x} \log y + y \cdot x^{y-1} + x^{x}(1 + \log x)]}{x \cdot y^{x-1} + x^{y} \log x}$.

(A) The given function is $x^{y} + y^{x} = 1$.
Let $u = x^{y}$ and $v = y^{x}$.
Then the equation becomes $u + v = 1$.
Differentiating with respect to $x$,we get $\frac{du}{dx} + \frac{dv}{dx} = 0$ $(1)$.
For $u = x^{y}$,taking log on both sides: $\log u = y \log x$.
Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx}$.
So,$\frac{du}{dx} = x^{y} \left( \frac{y}{x} + \log x \frac{dy}{dx} \right) = y x^{y-1} + x^{y} \log x \frac{dy}{dx}$ $(2)$.
For $v = y^{x}$,taking log on both sides: $\log v = x \log y$.
Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = x \cdot \frac{1}{y} \frac{dy}{dx} + \log y \cdot 1$.
So,$\frac{dv}{dx} = y^{x} \left( \frac{x}{y} \frac{dy}{dx} + \log y \right) = x y^{x-1} \frac{dy}{dx} + y^{x} \log y$ $(3)$.
Substituting $(2)$ and $(3)$ into $(1)$:
$y x^{y-1} + x^{y} \log x \frac{dy}{dx} + x y^{x-1} \frac{dy}{dx} + y^{x} \log y = 0$.
$\frac{dy}{dx} (x^{y} \log x + x y^{x-1}) = -(y x^{y-1} + y^{x} \log y)$.
Therefore,$\frac{dy}{dx} = -\frac{y x^{y-1} + y^{x} \log y}{x^{y} \log x + x y^{x-1}}$.

(A) Given the function $y^{x} = x^{y}$.
Taking the natural logarithm on both sides,we get:
$x \log y = y \log x$
Differentiating both sides with respect to $x$ using the product rule:
$\log y \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(\log y) = \log x \cdot \frac{d}{dx}(y) + y \cdot \frac{d}{dx}(\log x)$
$\log y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} = \log x \cdot \frac{dy}{dx} + y \cdot \frac{1}{x}$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{x}{y} \frac{dy}{dx} - \log x \frac{dy}{dx} = \frac{y}{x} - \log y$
$\left( \frac{x}{y} - \log x \right) \frac{dy}{dx} = \frac{y}{x} - \log y$
$\left( \frac{x - y \log x}{y} \right) \frac{dy}{dx} = \frac{y - x \log y}{x}$
Therefore,$\frac{dy}{dx} = \frac{y}{x} \left( \frac{y - x \log y}{x - y \log x} \right)$.

(A) Given the function $xy = e^{(x-y)}$.
Taking the natural logarithm on both sides:
$\ln(xy) = \ln(e^{(x-y)})$
$\ln x + \ln y = (x - y) \ln e$
Since $\ln e = 1$,we have:
$\ln x + \ln y = x - y$
Now,differentiate both sides with respect to $x$:
$\frac{d}{dx}(\ln x) + \frac{d}{dx}(\ln y) = \frac{d}{dx}(x) - \frac{dy}{dx}$
$\frac{1}{x} + \frac{1}{y} \frac{dy}{dx} = 1 - \frac{dy}{dx}$
Rearrange the terms to group $\frac{dy}{dx}$:
$\frac{1}{y} \frac{dy}{dx} + \frac{dy}{dx} = 1 - \frac{1}{x}$
$\frac{dy}{dx} \left(\frac{1+y}{y}\right) = \frac{x-1}{x}$
$\frac{dy}{dx} = \frac{y(x-1)}{x(y+1)}$

(A) Given equation: $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x^{\frac{2}{3}}) + \frac{d}{dx}(y^{\frac{2}{3}}) = \frac{d}{dx}(a^{\frac{2}{3}})$
Using the chain rule:
$\frac{2}{3}x^{-\frac{1}{3}} + \frac{2}{3}y^{-\frac{1}{3}} \cdot \frac{dy}{dx} = 0$
$\frac{2}{3}y^{-\frac{1}{3}} \cdot \frac{dy}{dx} = -\frac{2}{3}x^{-\frac{1}{3}}$
$\frac{dy}{dx} = -\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$
$\frac{dy}{dx} = -\left(\frac{y}{x}\right)^{\frac{1}{3}} = -\sqrt[3]{\frac{y}{x}}$

(A) Given equation: $x \sqrt{1+y} + y \sqrt{1+x} = 0$
Rearranging the terms: $x \sqrt{1+y} = -y \sqrt{1+x}$
Squaring both sides: $x^2(1+y) = y^2(1+x)$
Expanding: $x^2 + x^2y = y^2 + xy^2$
Rearranging: $x^2 - y^2 = xy^2 - x^2y$
Factoring: $(x-y)(x+y) = -xy(x-y)$
Since $x \neq y$ (as $x \sqrt{1+y} = -y \sqrt{1+x}$ implies $x$ and $y$ have opposite signs unless $x=y=0$),we can divide by $(x-y)$:
$x+y = -xy$
$y + xy = -x$
$y(1+x) = -x$
$y = -\frac{x}{1+x}$
Differentiating with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = -\left[ \frac{(1+x)(1) - x(1)}{(1+x)^2} \right]$
$\frac{dy}{dx} = -\left[ \frac{1+x-x}{(1+x)^2} \right] = -\frac{1}{(1+x)^2}$
Hence,proved.

Given that,$\cos y = x \cos (a+y)$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\cos y) = \frac{d}{dx}(x \cos (a+y))$
Using the chain rule and product rule:
$-\sin y \frac{dy}{dx} = \cos (a+y) \cdot (1) + x \cdot (-\sin (a+y)) \frac{dy}{dx}$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$x \sin (a+y) \frac{dy}{dx} - \sin y \frac{dy}{dx} = \cos (a+y)$
$\frac{dy}{dx} [x \sin (a+y) - \sin y] = \cos (a+y) \quad \dots(1)$
From the original equation,$x = \frac{\cos y}{\cos (a+y)}$. Substituting this into equation $(1)$:
$\frac{dy}{dx} \left[ \frac{\cos y}{\cos (a+y)} \sin (a+y) - \sin y \right] = \cos (a+y)$
$\frac{dy}{dx} \left[ \frac{\cos y \sin (a+y) - \sin y \cos (a+y)}{\cos (a+y)} \right] = \cos (a+y)$
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$\frac{dy}{dx} \left[ \frac{\sin(a+y-y)}{\cos (a+y)} \right] = \cos (a+y)$
$\frac{dy}{dx} \left[ \frac{\sin a}{\cos (a+y)} \right] = \cos (a+y)$
$\frac{dy}{dx} = \frac{\cos^2 (a+y)}{\sin a}$.
Hence,proved.

(B) Given the equation of the curve $y=(1+x)^{2y}+\cos^{2}(\sin^{-1} x)$.
At $x=0$,$y=(1+0)^{2y}+\cos^{2}(\sin^{-1} 0) = 1+1 = 2$.
So,we need to find the normal at the point $(0, 2)$.
Rewrite the equation as $y=e^{2y \ln(1+x)} + (1-x^2)$.
Differentiating with respect to $x$:
$y' = e^{2y \ln(1+x)} \left[ 2y \cdot \frac{1}{1+x} + \ln(1+x) \cdot 2y' \right] - 2x$.
Substituting $x=0$ and $y=2$:
$y' = e^{2(2) \ln(1)} \left[ 2(2) \cdot \frac{1}{1+0} + \ln(1) \cdot 2y' \right] - 2(0)$.
$y' = e^0 [4 + 0] - 0 = 4$.
Thus,the slope of the tangent $m_t = 4$.
The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{4}$.
The equation of the normal at $(0, 2)$ is $y - 2 = -\frac{1}{4}(x - 0)$.
$4y - 8 = -x$,which simplifies to $x + 4y = 8$.

(A) Given equation: $y^{2}+\ln(\cos^{2}x) = y$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
At $x=0$,$\cos^{2}(0) = 1$,so $\ln(1) = 0$. The equation becomes $y^{2} = y$,which implies $y(y-1) = 0$,so $y=0$ or $y=1$.
Differentiating with respect to $x$: $2yy^{\prime} + \frac{1}{\cos^{2}x} \cdot 2\cos x \cdot (-\sin x) = y^{\prime}$.
Simplifying: $2yy^{\prime} - 2\tan x = y^{\prime}$.
At $x=0$,for both $y=0$ and $y=1$,we get $2y(0) - 2(0) = y^{\prime}$,which means $y^{\prime}(0) = 0$.
Differentiating again: $2y y^{\prime \prime} + 2(y^{\prime})^{2} - 2\sec^{2}x = y^{\prime \prime}$.
At $x=0$ and $y^{\prime}(0)=0$: $2y y^{\prime \prime} + 0 - 2(1) = y^{\prime \prime}$.
If $y=0$,then $0 - 2 = y^{\prime \prime} \implies y^{\prime \prime}(0) = -2$.
If $y=1$,then $2y^{\prime \prime} - 2 = y^{\prime \prime} \implies y^{\prime \prime}(0) = 2$.
In both cases,$|y^{\prime \prime}(0)| = 2$.

(D) Given the limit $L = \lim_{t \rightarrow x} \frac{t^{2} f^{2}(x) - x^{2} f^{2}(t)}{t - x} = 0$.
Applying $L$'$H$ôpital's rule with respect to $t$:
$L = \lim_{t \rightarrow x} \frac{2t f^{2}(x) - x^{2} \cdot 2f(t) f'(t)}{1} = 0$.
Substituting $t = x$:
$2x f^{2}(x) - 2x^{2} f(x) f'(x) = 0$.
Dividing by $2x f(x)$ (since $x > 0$ and $f(x) > 0$):
$f(x) - x f'(x) = 0 \Rightarrow \frac{f'(x)}{f(x)} = \frac{1}{x}$.
Integrating both sides with respect to $x$:
$\int \frac{f'(x)}{f(x)} dx = \int \frac{1}{x} dx \Rightarrow \ln|f(x)| = \ln|x| + C$.
Since $f(x) > 0$ and $x > 0$,we have $f(x) = Cx$.
Using the condition $f(1) = e$:
$e = C(1) \Rightarrow C = e$.
Thus,$f(x) = ex$.
If $f(x) = 1$,then $ex = 1$,which implies $x = \frac{1}{e}$.

(D) Given $\log _{e}(x+y)=4 x y$. At $x=0$,$\log _{e}(y)=0$,which implies $y=1$.
Differentiating both sides with respect to $x$:
$\frac{1}{x+y} \left(1+\frac{d y}{d x}\right) = 4y + 4x \frac{d y}{d x}$.
At $x=0$ and $y=1$:
$\frac{1}{1} \left(1+\frac{d y}{d x}\right) = 4(1) + 4(0) \frac{d y}{d x} \Rightarrow 1+\frac{d y}{d x} = 4 \Rightarrow \frac{d y}{d x} = 3$.
Now,differentiate $1+\frac{d y}{d x} = (x+y)(4y + 4x \frac{d y}{d x})$:
$\frac{d^{2} y}{d x^{2}} = (1+\frac{d y}{d x})(4y + 4x \frac{d y}{d x}) + (x+y)(4 \frac{d y}{d x} + 4 \frac{d y}{d x} + 4x \frac{d^{2} y}{d x^{2}})$.
Substitute $x=0, y=1, \frac{d y}{d x}=3$:
$\frac{d^{2} y}{d x^{2}} = (1+3)(4(1) + 0) + (0+1)(4(3) + 4(3) + 0)$.
$\frac{d^{2} y}{d x^{2}} = (4)(4) + (1)(24) = 16 + 24 = 40$.

(D) Given the equation $\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \,d t=\int_{0}^{x} f(t) \,d t$ for $0 \leq x \leq 1$.
Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus:
$\sqrt{1-\left(f^{\prime}(x)\right)^{2}}=f(x)$
Squaring both sides:
$1-\left(f^{\prime}(x)\right)^{2}=f^{2}(x)$
$\left(f^{\prime}(x)\right)^{2} = 1 - f^{2}(x)$
$f^{\prime}(x) = \sqrt{1 - f^{2}(x)}$ (since $f$ is non-negative and $f(0)=0$ implies $f'(0)=1$)
Separating variables:
$\frac{f^{\prime}(x)}{\sqrt{1-f^{2}(x)}}=1$
Integrating both sides:
$\sin^{-1}(f(x)) = x + C$
Since $f(0)=0$,we have $\sin^{-1}(0) = 0 + C$,which gives $C=0$.
Thus,$f(x) = \sin(x)$.
Now,we evaluate the limit:
$\lim_{x \rightarrow 0} \frac{1}{x^{2}} \int_{0}^{x} \sin(t) \,dt = \lim_{x \rightarrow 0} \frac{[-\cos(t)]_{0}^{x}}{x^{2}} = \lim_{x \rightarrow 0} \frac{1 - \cos(x)}{x^{2}}$
Using the standard limit $\lim_{x \rightarrow 0} \frac{1 - \cos(x)}{x^{2}} = \frac{1}{2}$.

(C) Given the curve $C: (x^{2}+y^{2}-3)+(x^{2}-y^{2}-1)^{5}=0$.
Since $(\alpha, \alpha)$ lies on $C$,we substitute $x=\alpha$ and $y=\alpha$:
$(\alpha^{2}+\alpha^{2}-3)+(\alpha^{2}-\alpha^{2}-1)^{5}=0$
$(2\alpha^{2}-3)+(-1)^{5}=0$
$2\alpha^{2}-3-1=0 \Rightarrow 2\alpha^{2}=4 \Rightarrow \alpha^{2}=2$. Since $\alpha > 0$,$\alpha = \sqrt{2}$.
Differentiating the equation of the curve with respect to $x$:
$2x + 2yy^{\prime} + 5(x^{2}-y^{2}-1)^{4}(2x - 2yy^{\prime}) = 0$.
At $(\sqrt{2}, \sqrt{2})$:
$2\sqrt{2} + 2\sqrt{2}y^{\prime} + 5(-1)^{4}(2\sqrt{2} - 2\sqrt{2}y^{\prime}) = 0$
$2\sqrt{2} + 2\sqrt{2}y^{\prime} + 10\sqrt{2} - 10\sqrt{2}y^{\prime} = 0$
$12\sqrt{2} - 8\sqrt{2}y^{\prime} = 0 \Rightarrow y^{\prime} = \frac{12}{8} = \frac{3}{2}$.
Differentiating again:
$2 + 2(y^{\prime})^{2} + 2yy^{\prime\prime} + 5[4(x^{2}-y^{2}-1)^{3}(2x-2yy^{\prime})^{2} + (x^{2}-y^{2}-1)^{4}(2-2(y^{\prime})^{2}-2yy^{\prime\prime})] = 0$.
At $(\sqrt{2}, \sqrt{2})$ and $y^{\prime} = \frac{3}{2}$:
$2 + 2(\frac{9}{4}) + 2\sqrt{2}y^{\prime\prime} + 5[4(-1)^{3}(2\sqrt{2}-2\sqrt{2}(\frac{3}{2}))^{2} + (-1)^{4}(2-2(\frac{9}{4})-2\sqrt{2}y^{\prime\prime})] = 0$
$2 + \frac{9}{2} + 2\sqrt{2}y^{\prime\prime} + 5[-4(-\sqrt{2})^{2} + (2 - \frac{9}{2} - 2\sqrt{2}y^{\prime\prime})] = 0$
$\frac{13}{2} + 2\sqrt{2}y^{\prime\prime} + 5[-8 - \frac{5}{2} - 2\sqrt{2}y^{\prime\prime}] = 0$
$\frac{13}{2} + 2\sqrt{2}y^{\prime\prime} - 40 - \frac{25}{2} - 10\sqrt{2}y^{\prime\prime} = 0$
$-8\sqrt{2}y^{\prime\prime} = 40 + \frac{25-13}{2} = 40 + 6 = 46 \Rightarrow y^{\prime\prime} = -\frac{46}{8\sqrt{2}} = -\frac{23}{4\sqrt{2}}$.
Now,$3y^{\prime} - y^{3}y^{\prime\prime} = 3(\frac{3}{2}) - (\sqrt{2})^{3}(-\frac{23}{4\sqrt{2}}) = \frac{9}{2} - (2\sqrt{2})(-\frac{23}{4\sqrt{2}}) = \frac{9}{2} + \frac{23}{2} = \frac{32}{2} = 16$.

(B) Given the equation $2x^y + 3y^x = 20$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(2x^y) + \frac{d}{dx}(3y^x) = 0$.
Using the formula $\frac{d}{dx}(a^b) = a^b \frac{d}{dx}(b \ln a)$,we get:
$2x^y \left(\frac{y}{x} + \ln x \cdot \frac{dy}{dx}\right) + 3y^x \left(\frac{x}{y} \cdot \frac{dy}{dx} + \ln y\right) = 0$.
At the point $(2, 2)$,$x=2$ and $y=2$:
$2(2^2) \left(\frac{2}{2} + \ln 2 \cdot \frac{dy}{dx}\right) + 3(2^2) \left(\frac{2}{2} \cdot \frac{dy}{dx} + \ln 2\right) = 0$.
$8(1 + \ln 2 \cdot \frac{dy}{dx}) + 12(\frac{dy}{dx} + \ln 2) = 0$.
$8 + 8 \ln 2 \cdot \frac{dy}{dx} + 12 \frac{dy}{dx} + 12 \ln 2 = 0$.
$\frac{dy}{dx} (12 + 8 \ln 2) = -(8 + 12 \ln 2)$.
$\frac{dy}{dx} = -\frac{8 + 12 \ln 2}{12 + 8 \ln 2} = -\frac{2(4 + 6 \ln 2)}{4(3 + 2 \ln 2)} = -\frac{2 + 3 \ln 2}{3 + 2 \ln 2}$.
Since $3 \ln 2 = \ln 2^3 = \ln 8$ and $2 \ln 2 = \ln 2^2 = \ln 4$,we have:
$\frac{dy}{dx} = -\left(\frac{2 + \log_e 8}{3 + \log_e 4}\right)$.

(D) Given $f(x)=x^5+2x^3+3x+1$.
First,we find the derivative $f^{\prime}(x) = 5x^4+6x^2+3$.
We are given $g(f(x))=x$. Differentiating both sides with respect to $x$ using the chain rule,we get $g^{\prime}(f(x)) \cdot f^{\prime}(x) = 1$.
To find $g(7)$ and $g^{\prime}(7)$,we set $f(x)=7$:
$x^5+2x^3+3x+1=7 \Rightarrow x^5+2x^3+3x-6=0$.
By inspection,$x=1$ is a solution since $1+2+3-6=0$.
Thus,$f(1)=7$,which implies $g(7)=1$.
Now,substitute $x=1$ into the derivative equation $g^{\prime}(f(1)) \cdot f^{\prime}(1) = 1$:
$g^{\prime}(7) \cdot f^{\prime}(1) = 1$.
We calculate $f^{\prime}(1) = 5(1)^4+6(1)^2+3 = 5+6+3 = 14$.
Therefore,$g^{\prime}(7) = \frac{1}{f^{\prime}(1)} = \frac{1}{14}$.
Finally,$\frac{g(7)}{g^{\prime}(7)} = \frac{1}{1/14} = 14$.

(A, D, B) $1.$ Differentiating $y^3-3y+x=0$ with respect to $x$,we get $3y^2y^{\prime}-3y^{\prime}+1=0$,which implies $y^{\prime} = \frac{-1}{3(y^2-1)}$.
At $x = -10\sqrt{2}$,$y = 2\sqrt{2}$,so $y^{\prime} = \frac{-1}{3((2\sqrt{2})^2-1)} = \frac{-1}{3(8-1)} = -\frac{1}{21}$.
Differentiating $3y^2y^{\prime}-3y^{\prime}+1=0$ again,we get $6yy^{\prime 2} + 3y^2y^{\prime\prime} - 3y^{\prime\prime} = 0$.
$y^{\prime\prime}(3y^2-3) = -6yy^{\prime 2} \Rightarrow y^{\prime\prime} = \frac{-2yy^{\prime 2}}{y^2-1}$.
Substituting $y=2\sqrt{2}$ and $y^{\prime}=-\frac{1}{21}$,we get $f^{\prime\prime}(-10\sqrt{2}) = \frac{-2(2\sqrt{2})(-1/21)^2}{8-1} = \frac{-4\sqrt{2}}{7 \times 441} = -\frac{4\sqrt{2}}{7^3 \times 3^2}$.
$2.$ The area is $\int_a^b |f(x)| dx$. Since $f(x) < -2$ for $x < -2$,the area is $-\int_a^b f(x) dx$.
Using integration by parts: $\int f(x) dx = xf(x) - \int xf^{\prime}(x) dx$.
Since $f^{\prime}(x) = \frac{-1}{3(f(x)^2-1)}$,the integral is $bf(b)-af(a) - \int_a^b x \left(\frac{-1}{3(f(x)^2-1)}\right) dx = bf(b)-af(a) + \int_a^b \frac{x}{3(f(x)^2-1)} dx$.
The area is $-\int_a^b f(x) dx = -bf(b)+af(a) - \int_a^b \frac{x}{3(f(x)^2-1)} dx$.
$3.$ $\int_{-1}^1 g^{\prime}(x) dx = g(1) - g(-1)$.
Since $g(x)^3 - 3g(x) + x = 0$,$g(-x)^3 - 3g(-x) - x = 0$. Let $h(x) = -g(-x)$,then $(-h(x))^3 - 3(-h(x)) - x = 0 \Rightarrow -h(x)^3 + 3h(x) - x = 0 \Rightarrow h(x)^3 - 3h(x) + x = 0$.
Thus $g(x) = -g(-x)$,so $g$ is an odd function. $g(-1) = -g(1)$.
Therefore,$g(1) - g(-1) = g(1) - (-g(1)) = 2g(1)$.

(D) Given $f^{\prime}(x) = e^{f(x)} e^{-g(x)} g^{\prime}(x)$.
Dividing by $e^{f(x)}$,we get $e^{-f(x)} f^{\prime}(x) = e^{-g(x)} g^{\prime}(x)$.
Integrating both sides with respect to $x$,we get $\int e^{-f(x)} f^{\prime}(x) dx = \int e^{-g(x)} g^{\prime}(x) dx$.
This yields $-e^{-f(x)} = -e^{-g(x)} + C$,or $e^{-g(x)} - e^{-f(x)} = C$.
Using the condition $f(1) = 1$ and $g(2) = 1$,we evaluate the constant $C$ at different points.
Since $e^{-g(x)} - e^{-f(x)} = C$ holds for all $x$,we have $e^{-g(1)} - e^{-f(1)} = e^{-g(2)} - e^{-f(2)}$.
Substituting the given values $f(1) = 1$ and $g(2) = 1$,we get $e^{-g(1)} - e^{-1} = e^{-1} - e^{-f(2)}$.
Rearranging gives $e^{-f(2)} + e^{-g(1)} = 2e^{-1} = \frac{2}{e}$.
Since $e^{-f(2)} > 0$ and $e^{-g(1)} > 0$,we must have $e^{-f(2)} < \frac{2}{e}$ and $e^{-g(1)} < \frac{2}{e}$.
Taking the natural logarithm on both sides,$-f(2) < \ln(2) - 1$,which implies $f(2) > 1 - \ln(2)$.
Similarly,$-g(1) < \ln(2) - 1$,which implies $g(1) > 1 - \ln(2)$.
Thus,statements $(B)$ and $(C)$ are true.

(A) Given $\int_0^1 f(\lambda x) d\lambda = a f(x)$.
Let $\lambda x = t$,then $d\lambda = \frac{1}{x} dt$.
Substituting these into the integral: $\frac{1}{x} \int_0^x f(t) dt = a f(x)$,which implies $\int_0^x f(t) dt = a x f(x)$.
Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus: $f(x) = a(f(x) + x f^{\prime}(x))$.
Rearranging gives $(1 - a) f(x) = a x f^{\prime}(x)$,so $\frac{f^{\prime}(x)}{f(x)} = \frac{1 - a}{a} \cdot \frac{1}{x}$.
Integrating both sides: $\ln|f(x)| = \frac{1 - a}{a} \ln x + C$.
Since $f(1) = 1$,we get $C = 0$,so $f(x) = x^{\frac{1-a}{a}}$.
Given $f(16) = \frac{1}{8}$,we have $16^{\frac{1-a}{a}} = 2^{-3}$.
Since $16 = 2^4$,we have $2^{4 \cdot \frac{1-a}{a}} = 2^{-3}$,so $\frac{4(1-a)}{a} = -3$.
$4 - 4a = -3a \Rightarrow a = 4$.
Thus,$f(x) = x^{\frac{1-4}{4}} = x^{-3/4}$.
Then $f^{\prime}(x) = -\frac{3}{4} x^{-7/4}$.
$f^{\prime}\left(\frac{1}{16}\right) = -\frac{3}{4} \left(2^{-4}\right)^{-7/4} = -\frac{3}{4} \cdot 2^7 = -\frac{3}{4} \cdot 128 = -3 \cdot 32 = -96$.
Therefore,$16 - f^{\prime}\left(\frac{1}{16}\right) = 16 - (-96) = 112$.

(B) Given $f^{\prime\prime}(x) = f(x)$. Multiplying both sides by $f^{\prime}(x)$,we get $f^{\prime}(x)f^{\prime\prime}(x) = f(x)f^{\prime}(x)$.
Integrating both sides with respect to $x$,we have $\frac{1}{2}(f^{\prime}(x))^2 = \frac{1}{2}(f(x))^2 + C$.
Using the initial conditions $f(0) = 0$ and $f^{\prime}(0) = 3$,we get $\frac{1}{2}(3)^2 = \frac{1}{2}(0)^2 + C$,so $C = \frac{9}{2}$.
Thus,$(f^{\prime}(x))^2 = (f(x))^2 + 9$. Since $f^{\prime}(x) \geq 0$,we have $f^{\prime}(x) = \sqrt{(f(x))^2 + 9}$.
Separating variables,$\int \frac{df}{\sqrt{f^2 + 9}} = \int dx$,which gives $\ln|f(x) + \sqrt{(f(x))^2 + 9}| = x + C_1$.
Using $f(0) = 0$,we get $\ln|0 + \sqrt{0 + 9}| = 0 + C_1$,so $C_1 = \ln 3$.
Therefore,$f(x) + \sqrt{(f(x))^2 + 9} = 3e^x$.
Let $y = f(x)$. Then $\sqrt{y^2 + 9} = 3e^x - y$. Squaring both sides,$y^2 + 9 = 9e^{2x} - 6ye^x + y^2$,which simplifies to $6ye^x = 9e^{2x} - 9$.
So,$f(x) = \frac{9(e^{2x} - 1)}{6e^x} = \frac{3}{2}(e^x - e^{-x}) = 3\sinh(x)$.
At $x = \ln 3$,$f(\ln 3) = \frac{3}{2}(3 - \frac{1}{3}) = \frac{3}{2}(\frac{8}{3}) = 4$.
Thus,$9f(\ln 3) = 9 \times 4 = 36$.

(A) The given equation of the curve is $x^4-2xy^2+y^2+3x-3y=0 \dots (i)$.
To find the angle at which the curve cuts the $X$-axis,we need to find the slope of the tangent at the point $(0,0)$.
Differentiating equation $(i)$ with respect to $x$,we get:
$4x^3 - 2(y^2 + x \cdot 2y \frac{dy}{dx}) + 2y \frac{dy}{dx} + 3 - 3 \frac{dy}{dx} = 0$.
Simplifying the expression:
$4x^3 - 2y^2 - 4xy \frac{dy}{dx} + 2y \frac{dy}{dx} + 3 - 3 \frac{dy}{dx} = 0$.
Now,substitute $(x,y) = (0,0)$ into the derivative equation:
$4(0)^3 - 2(0)^2 - 4(0)(0) \frac{dy}{dx} + 2(0) \frac{dy}{dx} + 3 - 3 \frac{dy}{dx} = 0$.
$0 - 0 - 0 + 0 + 3 - 3 \frac{dy}{dx} = 0$.
$3 = 3 \frac{dy}{dx} \implies \frac{dy}{dx} = 1$.
Since the slope $m = \tan \theta = 1$,we have $\theta = \tan^{-1}(1) = \frac{\pi}{4}$.

(A) Given the curve is $xy + ax + by = 0$.
Since the point $(1, 1)$ lies on the curve,we substitute $x = 1$ and $y = 1$:
$1(1) + a(1) + b(1) = 0 \implies a + b = -1$ ... $(i)$
Now,differentiate the equation $xy + ax + by = 0$ with respect to $x$:
$x \frac{dy}{dx} + y + a + b \frac{dy}{dx} = 0$
$(x + b) \frac{dy}{dx} = -(y + a)$
$\frac{dy}{dx} = -\frac{y + a}{x + b}$
Given that the slope at $(1, 1)$ is $2$:
$2 = -\frac{1 + a}{1 + b}$
$2(1 + b) = -(1 + a)$
$2 + 2b = -1 - a$
$a + 2b = -3$ ... $(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(a + 2b) - (a + b) = -3 - (-1)$
$b = -2$
Substituting $b = -2$ into equation $(i)$:
$a - 2 = -1 \implies a = 1$
Therefore,$a - b = 1 - (-2) = 3$.

(B) Given equation: $x^k + y^k = a^k$
Differentiating with respect to $x$:
$k x^{k-1} + k y^{k-1} \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{k x^{k-1}}{k y^{k-1}} = -\frac{x^{k-1}}{y^{k-1}} = -(\frac{x}{y})^{k-1}$
$\frac{dy}{dx} = -(\frac{y}{x})^{-(k-1)} = -(\frac{y}{x})^{1-k}$
So,$\frac{dy}{dx} + (\frac{y}{x})^{1-k} = 0$
Comparing this with the given equation $\frac{dy}{dx} + (\frac{y}{x})^{\frac{1}{3}} = 0$:
$1 - k = \frac{1}{3}$
$k = 1 - \frac{1}{3} = \frac{2}{3}$

(B) Given $\frac{x}{x-y} = \log a - \log(x-y)$.
Rearranging,we get $\log(x-y) + \frac{x}{x-y} = \log a$.
Differentiating both sides with respect to $x$:
$\frac{1}{x-y} \left(1 - \frac{dy}{dx}\right) + \frac{(x-y)(1) - x(1 - \frac{dy}{dx})}{(x-y)^2} = 0$.
Multiply by $(x-y)^2$:
$(x-y)(1 - \frac{dy}{dx}) + x - y - x + x \frac{dy}{dx} = 0$.
$x - y - (x-y) \frac{dy}{dx} - y + x \frac{dy}{dx} = 0$.
$x - 2y + \frac{dy}{dx} (x - (x-y)) = 0$.
$x - 2y + y \frac{dy}{dx} = 0$.
$y \frac{dy}{dx} = 2y - x$.
Therefore,$\frac{dy}{dx} = \frac{2y - x}{y}$.