Differentiation of implicit function Questions in English

(C) Given: $3 f(\cos x) + 2 f(\sin x) = 5 x$ ...$(i)$
Replace $x$ with $(\frac{\pi}{2} - x)$:
$3 f(\sin x) + 2 f(\cos x) = 5(\frac{\pi}{2} - x)$ ...$(ii)$
Multiply $(i)$ by $3$ and $(ii)$ by $2$:
$9 f(\cos x) + 6 f(\sin x) = 15 x$
$4 f(\cos x) + 6 f(\sin x) = 10(\frac{\pi}{2} - x) = 5\pi - 10 x$
Subtracting the equations:
$5 f(\cos x) = 25 x - 5\pi \Rightarrow f(\cos x) = 5 x - \pi$
Differentiating with respect to $x$:
$f^{\prime}(\cos x) \cdot (-\sin x) = 5 \Rightarrow f^{\prime}(\cos x) = \frac{-5}{\sin x}$
Similarly,from $(i)$ and $(ii)$,we find $f(\sin x) = \frac{5\pi}{2} - 5x$:
$f^{\prime}(\sin x) \cdot (\cos x) = -5 \Rightarrow f^{\prime}(\sin x) = \frac{-5}{\cos x}$
Therefore,$f^{\prime}(\cos x) + f^{\prime}(\sin x) = \frac{-5}{\sin x} - \frac{5}{\cos x}$.

(C) Given the equation: $\frac{y}{x} \cos^4 \alpha + \frac{x}{y} \sin^4 \alpha = 2 \sin^2 \alpha \cos^2 \alpha$
Multiply both sides by $xy$:
$y^2 \cos^4 \alpha + x^2 \sin^4 \alpha = 2xy \sin^2 \alpha \cos^2 \alpha$
Rearrange the terms:
$x^2 \sin^4 \alpha - 2xy \sin^2 \alpha \cos^2 \alpha + y^2 \cos^4 \alpha = 0$
This is a perfect square:
$(x \sin^2 \alpha - y \cos^2 \alpha)^2 = 0$
Taking the square root on both sides:
$x \sin^2 \alpha - y \cos^2 \alpha = 0$
$y \cos^2 \alpha = x \sin^2 \alpha$
$y = x \frac{\sin^2 \alpha}{\cos^2 \alpha}$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{\sin^2 \alpha}{\cos^2 \alpha}$

(C) Given equations are:
$8 f(x)+6 f\left(\frac{1}{x}\right)=x+5$ --- $(i)$
$y=x^2 f(x)$ --- $(ii)$
Replace $x$ with $\frac{1}{x}$ in equation $(i)$:
$8 f\left(\frac{1}{x}\right)+6 f(x)=\frac{1}{x}+5$ --- $(iii)$
To eliminate $f\left(\frac{1}{x}\right)$,multiply $(i)$ by $4$ and $(iii)$ by $3$:
$32 f(x)+24 f\left(\frac{1}{x}\right)=4x+20$
$18 f(x)+24 f\left(\frac{1}{x}\right)=\frac{3}{x}+15$
Subtracting the second from the first:
$14 f(x)=4x-\frac{3}{x}+5$
Multiply by $x^2$ to get $y$:
$14 x^2 f(x)=4x^3-3x+5x^2$
$14 y=4x^3+5x^2-3x$
Differentiating with respect to $x$:
$14 \frac{d y}{d x}=12x^2+10x-3$
At $x=-1$:
$14 \left(\frac{d y}{d x}\right)_{x=-1}=12(-1)^2+10(-1)-3 = 12-10-3 = -1$
$\frac{d y}{d x} = -\frac{1}{14}$

(A) Given equation is $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(2x^2) - \frac{d}{dx}(3xy) + \frac{d}{dx}(y^2) + \frac{d}{dx}(x) + \frac{d}{dx}(2y) - \frac{d}{dx}(8) = 0$
$4x - 3(y + x \frac{dy}{dx}) + 2y \frac{dy}{dx} + 1 + 2 \frac{dy}{dx} = 0$
$4x - 3y - 3x \frac{dy}{dx} + 2y \frac{dy}{dx} + 1 + 2 \frac{dy}{dx} = 0$
Grouping the $\frac{dy}{dx}$ terms:
$(2y - 3x + 2) \frac{dy}{dx} = 3y - 4x - 1$
Therefore,$\frac{dy}{dx} = \frac{3y - 4x - 1}{2y - 3x + 2}$.
Thus,option $A$ is correct.

(C) Let $x^3 = \sin \theta$ and $y^3 = \sin \phi$. Then the equation becomes $\cos \theta + \cos \phi = a(\sin \theta - \sin \phi)$.
Using trigonometric identities,$2 \cos \frac{\theta+\phi}{2} \cos \frac{\theta-\phi}{2} = a \cdot 2 \cos \frac{\theta+\phi}{2} \sin \frac{\theta-\phi}{2}$.
This implies $\cot \frac{\theta-\phi}{2} = a$,which is a constant.
Thus,$\frac{\theta-\phi}{2} = \text{constant} \Rightarrow \theta - \phi = C$.
Substituting back,$\arcsin(x^3) - \arcsin(y^3) = C$.
Differentiating with respect to $x$:
$\frac{3x^2}{\sqrt{1-x^6}} - \frac{3y^2}{\sqrt{1-y^6}} \frac{dy}{dx} = 0$.
Rearranging the terms,we get $\frac{3y^2}{\sqrt{1-y^6}} \frac{dy}{dx} = \frac{3x^2}{\sqrt{1-x^6}}$.
Therefore,$y^2 \frac{dy}{dx} = x^2 \sqrt{\frac{1-y^6}{1-x^6}}$.

(A) Given equation is $\sqrt{x(1-y)} + \sqrt{y(1-x)} = 1$.
Let $x = \sin^2 \theta$ and $y = \sin^2 \phi$.
Then $\sqrt{\sin^2 \theta (1-\sin^2 \phi)} + \sqrt{\sin^2 \phi (1-\sin^2 \theta)} = 1$.
$\sin \theta \cos \phi + \sin \phi \cos \theta = 1$.
$\sin(\theta + \phi) = 1$.
$\theta + \phi = \frac{\pi}{2}$.
$\phi = \frac{\pi}{2} - \theta$.
Since $y = \sin^2 \phi$,we have $y = \sin^2(\frac{\pi}{2} - \theta) = \cos^2 \theta$.
Thus $x = \sin^2 \theta$ and $y = \cos^2 \theta$.
Adding these,$x+y = \sin^2 \theta + \cos^2 \theta = 1$,so $y = 1-x$.
Differentiating with respect to $x$,$\frac{dy}{dx} = -1$.
Alternatively,checking the options: $\frac{dy}{dx} = -\sqrt{\frac{y(1-y)}{x(1-x)}} = -\sqrt{\frac{(1-x)x}{x(1-x)}} = -\sqrt{1} = -1$.
Thus,option $A$ is correct.

(B) Given the equation: $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2$.
Differentiating both sides with respect to $x$:
$(a+\sqrt{2} b \cos x) \cdot (\sqrt{2} b \sin y \frac{dy}{dx}) + (a-\sqrt{2} b \cos y) \cdot (-\sqrt{2} b \sin x) = 0$.
At the point $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$,we have $\cos x = \cos y = \frac{1}{\sqrt{2}}$ and $\sin x = \sin y = \frac{1}{\sqrt{2}}$.
Substituting these values:
$(a+\sqrt{2} b \cdot \frac{1}{\sqrt{2}}) \cdot (\sqrt{2} b \cdot \frac{1}{\sqrt{2}} \cdot \frac{dy}{dx}) + (a-\sqrt{2} b \cdot \frac{1}{\sqrt{2}}) \cdot (-\sqrt{2} b \cdot \frac{1}{\sqrt{2}}) = 0$.
$(a+b) \cdot (b \frac{dy}{dx}) + (a-b) \cdot (-b) = 0$.
$(a+b) b \frac{dy}{dx} = b(a-b)$.
Since $b > 0$,we can divide by $b$:
$(a+b) \frac{dy}{dx} = a-b$.
Therefore,$\frac{dy}{dx} = \frac{a-b}{a+b}$.

(B) Given the curve equation is $xy + ax + by = 0$.
Since the point $(1, 1)$ lies on the curve,we substitute $x=1$ and $y=1$ into the equation:
$1(1) + a(1) + b(1) = 0 \implies 1 + a + b = 0 \implies a + b = -1$.
Now,differentiate the equation with respect to $x$:
$y + x \frac{dy}{dx} + a + b \frac{dy}{dx} = 0$.
Rearranging for $\frac{dy}{dx}$:
$\frac{dy}{dx}(x + b) = -(y + a) \implies \frac{dy}{dx} = -\frac{y + a}{x + b}$.
At the point $(1, 1)$,the slope of the tangent is $m = \tan(\tan^{-1}(2)) = 2$.
Substituting $(1, 1)$ into the derivative:
$2 = -\frac{1 + a}{1 + b} \implies 2(1 + b) = -(1 + a) \implies 2 + 2b = -1 - a \implies a + 2b = -3$.
We have a system of two linear equations:
$1) a + b = -1$
$2) a + 2b = -3$
Subtracting $(1)$ from $(2)$:
$(a + 2b) - (a + b) = -3 - (-1) \implies b = -2$.
Substituting $b = -2$ into $(1)$:
$a - 2 = -1 \implies a = 1$.
Finally,calculate $\frac{ab}{a+b}$:
$\frac{(1)(-2)}{1 + (-2)} = \frac{-2}{-1} = 2$.

(D) Given curve: $x^4 e^y + 2 \sqrt{y+1} = 3$.
Differentiating both sides with respect to $x$:
$x^4 e^y y' + 4x^3 e^y + \frac{2 y'}{2 \sqrt{y+1}} = 0$.
At the point $(1,0)$,substitute $x=1$ and $y=0$:
$(1)^4 e^0 y' + 4(1)^3 e^0 + \frac{y'}{\sqrt{0+1}} = 0$.
$y' + 4 + y' = 0 \Rightarrow 2y' = -4 \Rightarrow y' = -2$.
The equation of the tangent at $(1,0)$ with slope $m = -2$ is:
$y - 0 = -2(x - 1) \Rightarrow y = -2x + 2 \Rightarrow 2x + y = 2$.
Now,check which point satisfies the equation $2x + y = 2$:
For option $D$ $(-2, 6)$: $2(-2) + 6 = -4 + 6 = 2$.
Thus,the point $(-2, 6)$ lies on the tangent.

(A) The equation of the given curve is $3y^2 = 2ax^2 + 6b$ . . . $(i)$
Since the curve passes through the point $P(3, -1)$,we substitute $x = 3$ and $y = -1$ into equation $(i)$:
$3(-1)^2 = 2a(3)^2 + 6b$
$3 = 18a + 6b$
Dividing by $3$,we get $6a + 2b = 1$ . . . $(ii)$
Now,we differentiate equation $(i)$ with respect to $x$ to find the gradient:
$6y \frac{dy}{dx} = 4ax$
Given that the gradient at $P(3, -1)$ is $-1$,we substitute $\frac{dy}{dx} = -1$,$x = 3$,and $y = -1$:
$6(-1)(-1) = 4a(3)$
$6 = 12a$
$a = 1/2$
Substituting $a = 1/2$ into equation $(ii)$:
$6(1/2) + 2b = 1$
$3 + 2b = 1$
$2b = -2$
$b = -1$
Thus,the values are $a = 1/2$ and $b = -1$.

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

(C) Given $\tan y = \cot \left(\frac{\pi}{4} - x\right)$.
Differentiating both sides with respect to $x$:
$\sec^2 y \frac{dy}{dx} = -\operatorname{cosec}^2 \left(\frac{\pi}{4} - x\right) \cdot (-1)$
$\sec^2 y \frac{dy}{dx} = \operatorname{cosec}^2 \left(\frac{\pi}{4} - x\right)$
$\frac{dy}{dx} = \frac{\operatorname{cosec}^2 \left(\frac{\pi}{4} - x\right)}{\sec^2 y}$
Since $\sec^2 y = 1 + \tan^2 y$ and $\tan y = \cot \left(\frac{\pi}{4} - x\right)$:
$\frac{dy}{dx} = \frac{\operatorname{cosec}^2 \left(\frac{\pi}{4} - x\right)}{1 + \tan^2 \left(\frac{\pi}{4} - x\right)}$

(C) Given $\sec (\log _2 y^2) = \operatorname{cosec} (\log _2 x^2)$.
Using the property $\log a^b = b \log a$,we have $\sec (2 \log _2 y) = \operatorname{cosec} (2 \log _2 x)$.
We know that $\operatorname{cosec} \theta = \sec (\frac{\pi}{2} - \theta)$,so $\sec (2 \log _2 y) = \sec (\frac{\pi}{2} - 2 \log _2 x)$.
This implies $2 \log _2 y = \pm (\frac{\pi}{2} - 2 \log _2 x) + 2n\pi$. Taking the principal value,$2 \log _2 y + 2 \log _2 x = \frac{\pi}{2}$.
$2 \log _2 (xy) = \frac{\pi}{2} \Rightarrow \log _2 (xy) = \frac{\pi}{4}$.
$xy = 2^{\frac{\pi}{4}}$.
Since $2^{\frac{\pi}{4}}$ is a constant,differentiating both sides with respect to $x$ gives:
$x \frac{dy}{dx} + y = 0$.
Therefore,$\frac{dy}{dx} = -\frac{y}{x}$.

(D) Given $f(x) = x^2 + g'(1)x + g''(2)$. Let $g'(1) = a$ and $g''(2) = b$. Then $f(x) = x^2 + ax + b$.
$f'(x) = 2x + a$ and $f''(x) = 2$.
Given $g(x) = f(1)x^2 + xf'(x) + f''(x)$.
Substituting $f(1) = 1 + a + b$,$f'(x) = 2x + a$,and $f''(x) = 2$:
$g(x) = (1 + a + b)x^2 + x(2x + a) + 2 = (1 + a + b + 2)x^2 + ax + 2 = (3 + a + b)x^2 + ax + 2$.
Now,$g'(x) = 2(3 + a + b)x + a$ and $g''(x) = 2(3 + a + b)$.
Using $g'(1) = a$: $2(3 + a + b) + a = a \implies 6 + 2a + 2b = 0 \implies a + b = -3$.
Using $g''(2) = b$: $2(3 + a + b) = b \implies 6 + 2a + 2b = b \implies 2a + b = -6$.
Solving the system: $(2a + b) - (a + b) = -6 - (-3) \implies a = -3$.
Then $-3 + b = -3 \implies b = 0$.
Thus,$f(x) = x^2 - 3x$ and $g(x) = (3 - 3 + 0)x^2 - 3x + 2 = -3x + 2$.
$f(x) - g(x) = (x^2 - 3x) - (-3x + 2) = x^2 - 2$.

(A) Given equation is $x \sqrt{1+y} + y \sqrt{1+x} = 0$.
Rearranging the terms,we get $x \sqrt{1+y} = -y \sqrt{1+x}$.
Squaring both sides,we get $x^2(1+y) = y^2(1+x)$.
Expanding this,$x^2 + x^2y = y^2 + xy^2$.
Rearranging terms,$x^2 - y^2 + x^2y - xy^2 = 0$.
Factoring,$(x-y)(x+y) + xy(x-y) = 0$.
Since $x \neq y$ (as $x \sqrt{1+y} = -y \sqrt{1+x}$ implies $x$ and $y$ have opposite signs),we can divide by $(x-y)$ to get $x+y+xy = 0$.
Solving for $y$,$y(1+x) = -x$,so $y = \frac{-x}{1+x}$.
Differentiating with respect to $x$ using the quotient rule,$\frac{dy}{dx} = \frac{(1+x)(-1) - (-x)(1)}{(1+x)^2}$.
Simplifying,$\frac{dy}{dx} = \frac{-1-x+x}{(1+x)^2} = \frac{-1}{(1+x)^2}$.

(A) Given equation is $2^x+2^y=2^{x+y}$.
Dividing both sides by $2^{x+y}$,we get $\frac{2^x}{2^{x+y}}+\frac{2^y}{2^{x+y}}=1$.
This simplifies to $2^{-y}+2^{-x}=1$.
Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(2^{-y})+\frac{d}{dx}(2^{-x})=\frac{d}{dx}(1)$.
Using the chain rule,we have $2^{-y} \cdot \log 2 \cdot (-\frac{dy}{dx}) + 2^{-x} \cdot \log 2 \cdot (-1) = 0$.
Dividing by $-\log 2$,we get $2^{-y} \frac{dy}{dx} + 2^{-x} = 0$.
Thus,$\frac{dy}{dx} = -\frac{2^{-x}}{2^{-y}} = -2^{y-x}$.
From the original equation $2^x+2^y=2^{x+y}$,we have $2^x = 2^{x+y}-2^y = 2^y(2^x-1)$,so $2^{-x} = \frac{1}{2^y-1}$ is not direct. Let's use $2^{-x} = 1-2^{-y}$.
Then $\frac{dy}{dx} = -\frac{1-2^{-y}}{2^{-y}} = -(\frac{1}{2^{-y}}-1) = -(2^y-1) = 1-2^y$.

(D) Given equation is $\sqrt{\frac{y}{x}}+4 \sqrt{\frac{x}{y}}=4$.
Let $u = \sqrt{\frac{y}{x}}$. Then the equation becomes $u + \frac{4}{u} = 4$.
Multiplying by $u$,we get $u^2 - 4u + 4 = 0$,which is $(u-2)^2 = 0$.
Thus,$u = 2$,which implies $\sqrt{\frac{y}{x}} = 2$.
Squaring both sides,we get $\frac{y}{x} = 4$,or $y = 4x$.
Now,differentiating both sides with respect to $x$,we get $\frac{d y}{d x} = \frac{d}{d x}(4x) = 4$.
Therefore,$\frac{d y}{d x} = 4$.

(C) Given the equation: $\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}=2$
Multiplying both sides by $\sqrt{xy}$,we get: $y+x=2\sqrt{xy}$
Squaring both sides: $(x+y)^2 = (2\sqrt{xy})^2$
$x^2+y^2+2xy = 4xy$
$x^2+y^2-2xy = 0$
$(x-y)^2 = 0$
This implies $x-y=0$,or $y=x$
Differentiating both sides with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x) = 1$

(B) Given the equation $x^y = e^{x-y}$.
Taking the natural logarithm on both sides,we get:
$\ln(x^y) = \ln(e^{x-y})$
$y \ln x = x - y$
Rearranging the terms to isolate $y$:
$y \ln x + y = x$
$y(1 + \ln x) = x$
$y = \frac{x}{1 + \ln x}$
Now,differentiate $y$ with respect to $x$ using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$:
Let $u = x$ and $v = 1 + \ln x$. Then $u' = 1$ and $v' = \frac{1}{x}$.
$\frac{dy}{dx} = \frac{(1 + \ln x)(1) - x(\frac{1}{x})}{(1 + \ln x)^2}$
$\frac{dy}{dx} = \frac{1 + \ln x - 1}{(1 + \ln x)^2}$
$\frac{dy}{dx} = \frac{\ln x}{(1 + \ln x)^2}$
Thus,the correct option is $B$.

(B) Given $x = \sqrt{1-\tan y}$.
Squaring both sides,we get $x^2 = 1 - \tan y$.
Rearranging the terms,$\tan y = 1 - x^2$.
Now,differentiate both sides with respect to $x$:
$\frac{d}{dx}(\tan y) = \frac{d}{dx}(1 - x^2)$.
Using the chain rule,$\sec^2 y \cdot \frac{dy}{dx} = -2x$.
We know that $\sec^2 y = 1 + \tan^2 y$.
Substitute $\tan y = 1 - x^2$ into the expression:
$\sec^2 y = 1 + (1 - x^2)^2 = 1 + (1 - 2x^2 + x^4) = x^4 - 2x^2 + 2$.
Now,substitute this back into the derivative equation:
$(x^4 - 2x^2 + 2) \cdot \frac{dy}{dx} = -2x$.
Therefore,$\frac{dy}{dx} = -\frac{2x}{x^4 - 2x^2 + 2}$.

(C) Given the equation: $2x^2 - 3xy + 4y^2 + 2x - 3y + 4 = 0$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(2x^2) - \frac{d}{dx}(3xy) + \frac{d}{dx}(4y^2) + \frac{d}{dx}(2x) - \frac{d}{dx}(3y) + \frac{d}{dx}(4) = 0$.
$4x - 3(y + x \frac{dy}{dx}) + 8y \frac{dy}{dx} + 2 - 3 \frac{dy}{dx} = 0$.
$4x - 3y - 3x \frac{dy}{dx} + 8y \frac{dy}{dx} + 2 - 3 \frac{dy}{dx} = 0$.
Now,substitute the point $(3, 2)$ where $x = 3$ and $y = 2$:
$4(3) - 3(2) - 3(3) \frac{dy}{dx} + 8(2) \frac{dy}{dx} + 2 - 3 \frac{dy}{dx} = 0$.
$12 - 6 - 9 \frac{dy}{dx} + 16 \frac{dy}{dx} + 2 - 3 \frac{dy}{dx} = 0$.
$8 + 4 \frac{dy}{dx} = 0$.
$4 \frac{dy}{dx} = -8$.
$\frac{dy}{dx} = -2$.

(B) Given the equation: $2x^2 + 3xy - y^2 + 4x - 5y + 6 = 0$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(2x^2) + \frac{d}{dx}(3xy) - \frac{d}{dx}(y^2) + \frac{d}{dx}(4x) - \frac{d}{dx}(5y) + \frac{d}{dx}(6) = 0$.
$4x + 3(x \frac{dy}{dx} + y) - 2y \frac{dy}{dx} + 4 - 5 \frac{dy}{dx} = 0$.
Grouping the $\frac{dy}{dx}$ terms:
$(3x - 2y - 5) \frac{dy}{dx} = -(4x + 3y + 4)$.
$\frac{dy}{dx} = -\frac{4x + 3y + 4}{3x - 2y - 5}$.
Substituting $(x, y) = (1, -2)$:
$\frac{dy}{dx} = -\frac{4(1) + 3(-2) + 4}{3(1) - 2(-2) - 5} = -\frac{4 - 6 + 4}{3 + 4 - 5} = -\frac{2}{2} = -1$.

(A) Given the equation $x \cos (k+y)=\cos y$.
We can write $x = \frac{\cos y}{\cos (k+y)}$.
Differentiating both sides with respect to $y$:
$\frac{dx}{dy} = \frac{\cos(k+y) \cdot (-\sin y) - \cos y \cdot (-\sin(k+y))}{\cos^2(k+y)}$
$\frac{dx}{dy} = \frac{\sin(k+y)\cos y - \cos(k+y)\sin y}{\cos^2(k+y)}$
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we get:
$\frac{dx}{dy} = \frac{\sin(k+y-y)}{\cos^2(k+y)} = \frac{\sin k}{\cos^2(k+y)}$.
Therefore,$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{\cos^2(k+y)}{\sin k}$.
Now,substitute $y = \frac{\pi}{2}$:
$\frac{dy}{dx} = \frac{\cos^2(k+\frac{\pi}{2})}{\sin k} = \frac{(-\sin k)^2}{\sin k} = \frac{\sin^2 k}{\sin k} = \sin k$.
Thus,option $A$ is correct.

(D) Given the equation $x e^{xy} = y + \sin^2 x$.
At $x = 0$,we have $0 \cdot e^{0 \cdot y} = y + \sin^2(0)$,which implies $y = 0$.
Now,differentiate both sides with respect to $x$:
$\frac{d}{dx}(x e^{xy}) = \frac{dy}{dx} + \frac{d}{dx}(\sin^2 x)$.
Using the product rule and chain rule:
$x e^{xy} \left( x \frac{dy}{dx} + y \right) + e^{xy} = \frac{dy}{dx} + 2 \sin x \cos x$.
Substituting $x = 0$ and $y = 0$:
$0 \cdot e^0 (0 \cdot \frac{dy}{dx} + 0) + e^0 = \frac{dy}{dx} + 2 \sin(0) \cos(0)$.
$0 + 1 = \frac{dy}{dx} + 0$.
Therefore,$\frac{dy}{dx} = 1$ at $x = 0$.

(D) Given the equation: $y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+\ldots \infty}}}}$
We can rewrite the inner part as: $y=\sqrt{x+\sqrt{y+y}}$
Squaring both sides: $y^2=x+\sqrt{2y}$
Rearranging: $y^2-x=\sqrt{2y}$
Squaring again: $(y^2-x)^2=2y$
Expanding: $y^4-2xy^2+x^2=2y$
Rearranging to implicit form: $y^4-2xy^2-2y+x^2=0$
Differentiating with respect to $x$: $\frac{d}{dx}(y^4-2xy^2-2y+x^2) = 0$
$4y^3 \frac{dy}{dx} - (2y^2 + 4xy \frac{dy}{dx}) - 2 \frac{dy}{dx} + 2x = 0$
Grouping $\frac{dy}{dx}$ terms: $\frac{dy}{dx}(4y^3 - 4xy - 2) = 2y^2 - 2x$
Solving for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{2y^2-2x}{4y^3-4xy-2} = \frac{y^2-x}{2y^3-2xy-1}$

(B) Given equation: $(x^2-3x+2) e^{\frac{y}{x-1}} = x+2$.
Factorizing the quadratic term: $(x-1)(x-2) e^{\frac{y}{x-1}} = x+2$.
Taking natural logarithm on both sides: $\ln((x-1)(x-2)) + \frac{y}{x-1} = \ln(x+2)$.
Rearranging for $y$: $y = (x-1) [\ln(x+2) - \ln((x-1)(x-2))]$.
At $x=0$: $y = (0-1) [\ln(2) - \ln((-1)(-2))] = -1 [\ln(2) - \ln(2)] = 0$.
Now,differentiate the original equation with respect to $x$:
$(2x-3) e^{\frac{y}{x-1}} + (x^2-3x+2) e^{\frac{y}{x-1}} \cdot \frac{d}{dx}(\frac{y}{x-1}) = 1$.
Using quotient rule for $\frac{d}{dx}(\frac{y}{x-1}) = \frac{(x-1)y' - y}{(x-1)^2}$.
Substitute $x=0$ and $y=0$:
$(0-3) e^{\frac{0}{-1}} + (0-0+2) e^{\frac{0}{-1}} \cdot \frac{(-1)y' - 0}{(-1)^2} = 1$.
$-3(1) + 2(1) (-y') = 1$.
$-3 - 2y' = 1$.
$-2y' = 4$.
$y' = -2$.
Thus,$(\frac{dy}{dx})_{x=0} = -2$.

(D) Given the equation: $3^x y^x = x^{3y}$.
Taking the natural logarithm on both sides:
$\ln(3^x y^x) = \ln(x^{3y})$
$x \ln 3 + x \ln y = 3y \ln x$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x \ln 3) + \frac{d}{dx}(x \ln y) = \frac{d}{dx}(3y \ln x)$
$\ln 3 + (\ln y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx}) = 3 \cdot \frac{dy}{dx} \cdot \ln x + 3y \cdot \frac{1}{x}$.
At $x = 1$,we first find $y$ from the original equation:
$3^1 y^1 = 1^{3y} \implies 3y = 1 \implies y = \frac{1}{3}$.
Substituting $x = 1$ and $y = \frac{1}{3}$ into the differentiated equation:
$\ln 3 + \ln(\frac{1}{3}) + 1 \cdot \frac{1}{1/3} \cdot \frac{dy}{dx} = 3 \cdot \frac{dy}{dx} \cdot \ln 1 + 3(\frac{1}{3}) \cdot \frac{1}{1}$.
Since $\ln(\frac{1}{3}) = -\ln 3$ and $\ln 1 = 0$:
$\ln 3 - \ln 3 + 3 \frac{dy}{dx} = 0 + 1$.
$3 \frac{dy}{dx} = 1$.
$\frac{dy}{dx} = \frac{1}{3}$.

(D) Given: $x^2+y^2=t-\frac{1}{t}$ (Equation $1$)
Squaring both sides of Equation $1$:
$(x^2+y^2)^2 = (t-\frac{1}{t})^2$
$x^4+y^4+2x^2y^2 = t^2+\frac{1}{t^2}-2$
Given: $x^4+y^4=t^2+\frac{1}{t^2}$ (Equation $2$)
Substituting Equation $2$ into the expanded form:
$(t^2+\frac{1}{t^2})+2x^2y^2 = t^2+\frac{1}{t^2}-2$
$2x^2y^2 = -2$
$x^2y^2 = -1$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x^2y^2) = \frac{d}{dx}(-1)$
$x^2(2y \frac{dy}{dx}) + y^2(2x) = 0$
$2x^2y \frac{dy}{dx} = -2xy^2$
$\frac{dy}{dx} = \frac{-2xy^2}{2x^2y} = -\frac{y}{x}$

(A) Given $\log y = y^{\log x}$.
Taking $\log$ on both sides:
$\log(\log y) = \log(y^{\log x}) = \log x \cdot \log y$.
Differentiating both sides with respect to $x$:
$\frac{1}{\log y} \cdot \frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}(\log x) \cdot \log y + \log x \cdot \frac{d}{dx}(\log y)$.
$\frac{1}{y \log y} \cdot \frac{dy}{dx} = \frac{\log y}{x} + \frac{\log x}{y} \cdot \frac{dy}{dx}$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx} \left( \frac{1}{y \log y} - \frac{\log x}{y} \right) = \frac{\log y}{x}$.
$\frac{dy}{dx} \left( \frac{1 - \log x \log y}{y \log y} \right) = \frac{\log y}{x}$.
$\frac{dy}{dx} = \frac{\log y}{x} \cdot \frac{y \log y}{1 - \log x \log y}$.
$\frac{dy}{dx} = \frac{y(\log y)^2}{x(1 - \log x \log y)}$.

(A) Given equation: $y^{\cos x}=x^{\sin y}$
Taking logarithm on both sides: $\cos x \log y = \sin y \log x$
Differentiating both sides with respect to $x$ using the product rule:
$\frac{d}{dx}(\cos x \log y) = \frac{d}{dx}(\sin y \log x)$
$(\cos x) \cdot \frac{1}{y} \frac{dy}{dx} + (\log y) \cdot (-\sin x) = (\sin y) \cdot \frac{1}{x} + (\log x) \cdot (\cos y) \frac{dy}{dx}$
Rearranging the terms to group $\frac{dy}{dx}$:
$\frac{dy}{dx} \left( \frac{\cos x}{y} - \cos y \log x \right) = \frac{\sin y}{x} + \sin x \log y$
$\frac{dy}{dx} \left( \frac{\cos x - y \cos y \log x}{y} \right) = \frac{\sin y + x \sin x \log y}{x}$
$\frac{dy}{dx} = \frac{y(\sin y + x \sin x \log y)}{x(\cos x - y \cos y \log x)}$

(B) Given the equation: $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(2x^2) - \frac{d}{dx}(3xy) + \frac{d}{dx}(y^2) + \frac{d}{dx}(x) + \frac{d}{dx}(2y) - \frac{d}{dx}(8) = 0$.
$4x - 3(x \frac{dy}{dx} + y) + 2y \frac{dy}{dx} + 1 + 2 \frac{dy}{dx} = 0$.
Grouping the $\frac{dy}{dx}$ terms:
$\frac{dy}{dx}(2y - 3x + 2) + (4x - 3y + 1) = 0$.
This can be written as: $(2y - 3x + 2) dy + (4x - 3y + 1) dx = 0$.
Comparing this with $f(x, y) dy - g(x, y) dx = 0$,we get:
$f(x, y) = 2y - 3x + 2$ and $g(x, y) = -(4x - 3y + 1) = 3y - 4x - 1$.
Now,calculate $g(2, 2) = 3(2) - 4(2) - 1 = 6 - 8 - 1 = -3$.
Calculate $f(1, 1) = 2(1) - 3(1) + 2 = 2 - 3 + 2 = 1$.
Therefore,$\frac{g(2, 2)}{f(1, 1)} = \frac{-3}{1} = -3$.

(B) Given $y = \frac{ax + b}{cx + d}$.
To find $\frac{dx}{dy}$,we first express $x$ in terms of $y$:
$y(cx + d) = ax + b$
$cyx + yd = ax + b$
$cyx - ax = b - yd$
$x(cy - a) = b - yd$
$x = \frac{b - yd}{cy - a} = \frac{yd - b}{a - cy}$.
Now,differentiate $x$ with respect to $y$ using the quotient rule $\frac{d}{dy} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$:
$\frac{dx}{dy} = \frac{d(yd - b) \cdot (a - cy) - (yd - b) \cdot d(a - cy)}{dy} \cdot \frac{1}{(a - cy)^2}$
$\frac{dx}{dy} = \frac{d(a - cy) - (yd - b)(-c)}{(a - cy)^2}$
$\frac{dx}{dy} = \frac{ad - cdy + cdy - bc}{(a - cy)^2}$
$\frac{dx}{dy} = \frac{ad - bc}{(a - cy)^2}$.

(B) Given equations are $x^2+y^2=t+\frac{2}{t}$ and $x^4+y^4=t^2+\frac{4}{t^2}$.
Squaring the first equation: $(x^2+y^2)^2 = (t+\frac{2}{t})^2$.
$x^4+y^4+2x^2y^2 = t^2+\frac{4}{t^2}+4$.
Substituting the value of $x^4+y^4$ from the second equation: $(t^2+\frac{4}{t^2}) + 2x^2y^2 = t^2+\frac{4}{t^2}+4$.
This simplifies to $2x^2y^2 = 4$,which gives $x^2y^2 = 2$.
Thus,$y^2 = \frac{2}{x^2}$.
Differentiating both sides with respect to $x$: $2y \frac{dy}{dx} = -2 \cdot 2x^{-3} = -\frac{4}{x^3}$.
Multiplying both sides by $x^3$: $2x^3y \frac{dy}{dx} = -4$.
Therefore,$x^3y \frac{dy}{dx} = -2$.

(A) Given,$\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=k$
$\Rightarrow \frac{x^2-y^2}{x^2+y^2}=\cos k$
Let $\cos k = C$ (a constant).
Then,$x^2 - y^2 = C(x^2 + y^2)$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x^2 - y^2) = \frac{d}{dx}(C(x^2 + y^2))$
$2x - 2y \frac{dy}{dx} = C(2x + 2y \frac{dy}{dx})$
$x - y \frac{dy}{dx} = C(x + y \frac{dy}{dx})$
$x - Cx = Cy \frac{dy}{dx} + y \frac{dy}{dx}$
$x(1 - C) = y \frac{dy}{dx}(C + 1)$
$\frac{dy}{dx} = \frac{x(1 - C)}{y(1 + C)}$
Substituting $C = \frac{x^2-y^2}{x^2+y^2}$ back:
$\frac{dy}{dx} = \frac{x(1 - \frac{x^2-y^2}{x^2+y^2})}{y(1 + \frac{x^2-y^2}{x^2+y^2})} = \frac{x(\frac{x^2+y^2-x^2+y^2}{x^2+y^2})}{y(\frac{x^2+y^2+x^2-y^2}{x^2+y^2})} = \frac{x(2y^2)}{y(2x^2)} = \frac{xy^2}{yx^2} = \frac{y}{x}$

(C) Given the equation: $ax^2+2hxy+by^2=3$.
Differentiating with respect to $x$:
$2ax + 2h(y + x\frac{dy}{dx}) + 2by\frac{dy}{dx} = 0$.
Dividing by $2$: $ax + hy + hx\frac{dy}{dx} + by\frac{dy}{dx} = 0$.
Thus,$\frac{dy}{dx} = -\frac{ax+hy}{hx+by}$.
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = -\frac{(hx+by)(a+h\frac{dy}{dx}) - (ax+hy)(h+b\frac{dy}{dx})}{(hx+by)^2}$.
Substituting $\frac{dy}{dx} = -\frac{ax+hy}{hx+by}$ and simplifying using the original equation $ax^2+2hxy+by^2=3$,we get:
$\frac{d^2y}{dx^2} = \frac{3(h^2-ab)}{(hx+by)^3}$.

(B) Given $a f(x)+b f\left(\frac{1}{x}\right)=x+1$ $(i)$
Replacing $x$ with $\frac{1}{x}$,we get $a f\left(\frac{1}{x}\right)+b f(x)=\frac{1}{x}+1$ $(ii)$
Multiplying $(i)$ by $a$ and $(ii)$ by $b$,we get:
$a^2 f(x)+a b f\left(\frac{1}{x}\right)=a x+a$
$b^2 f(x)+a b f\left(\frac{1}{x}\right)=\frac{b}{x}+b$
Subtracting the two equations:
$(a^2-b^2) f(x)=a x-\frac{b}{x}+a-b$
$f(x)=\frac{a x}{a^2-b^2}-\frac{b}{x(a^2-b^2)}+\frac{1}{a+b}$
Now,$x^2 f(x)=\frac{a x^3}{a^2-b^2}-\frac{b x}{a^2-b^2}+\frac{x^2}{a+b}$
Differentiating with respect to $x$:
$\frac{d}{d x}\left(x^2 f(x)\right)=\frac{3 a x^2}{a^2-b^2}-\frac{b}{a^2-b^2}+\frac{2 x}{a+b}$
Comparing with $2 x^2+2 x+\frac{1}{3}$:
$\frac{3 a}{a^2-b^2}=2$ $(iii)$
$\frac{2}{a+b}=2 \Rightarrow a+b=1$ $(iv)$
$-\frac{b}{a^2-b^2}=\frac{1}{3}$ $(v)$
From $(iv)$,$a^2-b^2=(a-b)(a+b)=a-b$.
Substituting into $(iii)$ and $(v)$:
$\frac{3 a}{a-b}=2 \Rightarrow 3 a=2 a-2 b \Rightarrow a=-2 b$
$-\frac{b}{a-b}=\frac{1}{3} \Rightarrow -3 b=a-b \Rightarrow a=-2 b$
Since $a+b=1$ and $a=-2 b$,we have $-2 b+b=1 \Rightarrow b=-1$ and $a=2$.
Therefore,$a-b=2-(-1)=3$.

(D) Given,$f(x) = \frac{1}{x^3} \int_5^x (2u^2 - u f'(u)) du$
Multiplying by $x^3$,we get $x^3 f(x) = \int_5^x (2u^2 - u f'(u)) du$
Differentiating both sides with respect to $x$ using the Leibniz rule:
$x^3 f'(x) + 3x^2 f(x) = 2x^2 - x f'(x)$
Rearranging the terms to solve for $f'(x)$:
$x^3 f'(x) + x f'(x) = 2x^2 - 3x^2 f(x)$
$f'(x)(x^3 + x) = 2x^2 - 3x^2 f(x)$
$f'(x) = \frac{2x^2 - 3x^2 f(x)}{x^3 + x}$
At $x = 5$,we know $f(5) = \frac{1}{5^3} \int_5^5 (2u^2 - u f'(u)) du = 0$.
Substituting $x = 5$ and $f(5) = 0$ into the expression for $f'(x)$:
$f'(5) = \frac{2(5)^2 - 3(5)^2(0)}{5^3 + 5} = \frac{50}{125 + 5} = \frac{50}{130} = \frac{5}{13}$

(B) Given $f(x) = \prod_{r=1}^n \cos(rx)$. Taking the natural logarithm on both sides,we get $\ln|f(x)| = \sum_{r=1}^n \ln|\cos(rx)|$.
Differentiating both sides with respect to $x$,we have $\frac{f^{\prime}(x)}{f(x)} = \sum_{r=1}^n \frac{1}{\cos(rx)} \cdot (-\sin(rx) \cdot r) = -\sum_{r=1}^n r \tan(rx)$.
Multiplying both sides by $f(x)$,we get $f^{\prime}(x) = -f(x) \sum_{r=1}^n r \tan(rx)$.
Rearranging the terms,we obtain $f^{\prime}(x) + \sum_{r=1}^n (r \tan rx) f(x) = 0$.

(C) Given the function $f(x) = \tanh^{-1}(\sin x)$.
Let $y = \tanh^{-1}(\sin x)$,which implies $\tanh y = \sin x$.
Differentiating both sides with respect to $x$:
$\operatorname{sech}^2 y \cdot \frac{dy}{dx} = \cos x$.
Thus,$\frac{dy}{dx} = \frac{\cos x}{\operatorname{sech}^2 y}$.
Using the identity $\operatorname{sech}^2 y = 1 - \tanh^2 y$,we get:
$\frac{dy}{dx} = \frac{\cos x}{1 - \tanh^2 y}$.
Since $\tanh y = \sin x$,we have $\frac{dy}{dx} = \frac{\cos x}{1 - \sin^2 x} = \frac{\cos x}{\cos^2 x} = \sec x$.
At $x = \pi$,the slope is $\sec(\pi) = -1$.

(B) Given the equation: $\frac{2}{f} = \frac{1}{v} - \frac{1}{u}$ $(i)$
Taking the differential of both sides:
$-\frac{2}{f^2} df = -\frac{1}{v^2} dv + \frac{1}{u^2} du$
Given that the absolute errors in $u$ and $v$ are $p$,we have $du = p$ and $dv = p$.
Substituting these into the differential equation:
$-\frac{2}{f^2} df = -\frac{1}{v^2} p + \frac{1}{u^2} p$
$-\frac{2}{f^2} df = -p \left( \frac{1}{v^2} - \frac{1}{u^2} \right)$
Using the identity $a^2 - b^2 = (a - b)(a + b)$:
$-\frac{2}{f^2} df = -p \left( \frac{1}{v} - \frac{1}{u} \right) \left( \frac{1}{v} + \frac{1}{u} \right)$
Since $\frac{1}{v} - \frac{1}{u} = \frac{2}{f}$ from equation $(i)$:
$-\frac{2}{f^2} df = -p \left( \frac{2}{f} \right) \left( \frac{1}{v} + \frac{1}{u} \right)$
Dividing both sides by $-\frac{2}{f}$:
$\frac{df}{f} = p \left( \frac{1}{v} + \frac{1}{u} \right)$
Thus,the relative error in $f$ is $p \left( \frac{1}{u} + \frac{1}{v} \right)$.

(C) Given that,$f(x) = \frac{1}{x^2} \int_3^x (2t - 3f'(t)) dt$.
Multiplying both sides by $x^2$,we get $x^2 f(x) = \int_3^x (2t - 3f'(t)) dt$.
Differentiating both sides with respect to $x$ using the product rule on the left and the Fundamental Theorem of Calculus on the right:
$x^2 f'(x) + 2x f(x) = 2x - 3f'(x)$.
Rearranging the terms: $f'(x)(x^2 + 3) = 2x - 2x f(x)$.
At $x = 3$,note that $f(3) = \frac{1}{3^2} \int_3^3 (2t - 3f'(t)) dt = 0$.
Substituting $x = 3$ into the differentiated equation:
$f'(3)(3^2 + 3) = 2(3) - 2(3) f(3)$.
$f'(3)(9 + 3) = 6 - 6(0)$.
$12 f'(3) = 6$.
$f'(3) = \frac{6}{12} = \frac{1}{2}$.