Differentiation of implicit function Questions in English

(A) Given equation: $\log(x+y) = 2xy$ $(i)$
Differentiating both sides with respect to $x$:
$\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) = 2y + 2x \frac{dy}{dx}$ $(ii)$
At $x=0$,from $(i)$: $\log(0+y) = 2(0)y \implies \log(y) = 0 \implies y = e^0 = 1$.
Substituting $x=0$ and $y=1$ into $(ii)$:
$\frac{1}{0+1} \left(1 + \frac{dy}{dx}\right) = 2(1) + 2(0) \frac{dy}{dx}$
$1 + \frac{dy}{dx} = 2$
$\frac{dy}{dx} = 2 - 1 = 1$
Thus,the value of $\frac{dy}{dx}$ at $x=0$ is $1$.

(B) We know that for any real number $z$,$\tan^{-1}(z) + \cot^{-1}(z) = \frac{\pi}{2}$.
Given the equation $xy = \tan^{-1}(xy) + \cot^{-1}(xy)$,we can substitute $z = xy$ to get $xy = \frac{\pi}{2}$.
Now,differentiate both sides with respect to $x$ using the product rule:
$\frac{d}{dx}(xy) = \frac{d}{dx}(\frac{\pi}{2})$
$x \frac{dy}{dx} + y(1) = 0$
$x \frac{dy}{dx} = -y$
$\frac{dy}{dx} = -\frac{y}{x}$

(D) Given $a(4+x^2)=x$,we have $a = \frac{x}{4+x^2}$.
At $x=1$,$a = \frac{1}{4+1^2} = \frac{1}{5}$.
Differentiating $a(4+x^2)=x$ with respect to $x$ using the product rule:
$\frac{da}{dx}(4+x^2) + a(2x) = 1$.
Substituting $x=1$ and $a=\frac{1}{5}$:
$\frac{da}{dx}(4+1) + \frac{1}{5}(2(1)) = 1 \implies 5\frac{da}{dx} + \frac{2}{5} = 1 \implies 5\frac{da}{dx} = \frac{3}{5} \implies \frac{da}{dx} = \frac{3}{25}$.
Given $y = x^3 + a^2$,differentiating with respect to $x$:
$\frac{dy}{dx} = 3x^2 + 2a\frac{da}{dx}$.
At $x=1$,$a=\frac{1}{5}$ and $\frac{da}{dx}=\frac{3}{25}$:
$\frac{dy}{dx} = 3(1)^2 + 2(\frac{1}{5})(\frac{3}{25}) = 3 + \frac{6}{125} = \frac{375+6}{125} = \frac{381}{125}$.

(C) Let $y = \log \left[f\left(e^x+2 x\right)\right]$.
Using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{f\left(e^x+2 x\right)} \cdot f^{\prime}\left(e^x+2 x\right) \cdot \frac{d}{dx}(e^x+2 x)$.
$\frac{dy}{dx} = \frac{f^{\prime}\left(e^x+2 x\right) \cdot (e^x+2)}{f\left(e^x+2 x\right)}$.
Now,evaluate at $x=0$:
At $x=0$,the argument of $f$ is $e^0 + 2(0) = 1 + 0 = 1$.
So,$\left. \frac{dy}{dx} \right|_{x=0} = \frac{f^{\prime}(1) \cdot (e^0+2)}{f(1)}$.
Given $f(1)=3$ and $f^{\prime}(1)=2$:
$\left. \frac{dy}{dx} \right|_{x=0} = \frac{2 \cdot (1+2)}{3} = \frac{2 \cdot 3}{3} = 2$.

(D) Given the equation $x = e^{\tan^{-1}\left(\frac{y-x^2}{x^2}\right)}$.
Taking the natural logarithm on both sides,we get $\ln(x) = \tan^{-1}\left(\frac{y-x^2}{x^2}\right)$.
This implies $\tan(\ln x) = \frac{y-x^2}{x^2}$.
Rearranging the terms,we have $y = x^2 \tan(\ln x) + x^2$.
Differentiating both sides with respect to $x$ using the product rule and chain rule:
$\frac{dy}{dx} = \frac{d}{dx}[x^2 \tan(\ln x)] + \frac{d}{dx}[x^2]$
$\frac{dy}{dx} = [2x \tan(\ln x) + x^2 \cdot \sec^2(\ln x) \cdot \frac{1}{x}] + 2x$
$\frac{dy}{dx} = 2x \tan(\ln x) + x \sec^2(\ln x) + 2x$.
Now,evaluate at $x = 1$:
$\frac{dy}{dx} \Big|_{x=1} = 2(1) \tan(\ln 1) + 1 \sec^2(\ln 1) + 2(1)$
Since $\ln 1 = 0$ and $\tan 0 = 0$ and $\sec 0 = 1$:
$\frac{dy}{dx} \Big|_{x=1} = 2(0) + 1(1)^2 + 2 = 0 + 1 + 2 = 3$.

(C) Given $\tan y = \frac{x \sin \alpha}{1-x \cos \alpha}$.
Taking the derivative with respect to $x$ on both sides:
$\frac{d}{dx}(\tan y) = \frac{d}{dx}\left(\frac{x \sin \alpha}{1-x \cos \alpha}\right)$
$\sec^2 y \cdot \frac{dy}{dx} = \frac{(1-x \cos \alpha)(\sin \alpha) - (x \sin \alpha)(-\cos \alpha)}{(1-x \cos \alpha)^2}$
$\sec^2 y \cdot \frac{dy}{dx} = \frac{\sin \alpha - x \sin \alpha \cos \alpha + x \sin \alpha \cos \alpha}{(1-x \cos \alpha)^2} = \frac{\sin \alpha}{(1-x \cos \alpha)^2}$
Since $\tan y = \frac{x \sin \alpha}{1-x \cos \alpha}$,we have $\sec^2 y = 1 + \tan^2 y = 1 + \frac{x^2 \sin^2 \alpha}{(1-x \cos \alpha)^2} = \frac{(1-x \cos \alpha)^2 + x^2 \sin^2 \alpha}{(1-x \cos \alpha)^2} = \frac{1 - 2x \cos \alpha + x^2 \cos^2 \alpha + x^2 \sin^2 \alpha}{(1-x \cos \alpha)^2} = \frac{1 - 2x \cos \alpha + x^2}{(1-x \cos \alpha)^2}$.
Substituting this back:
$\frac{1 - 2x \cos \alpha + x^2}{(1-x \cos \alpha)^2} \cdot \frac{dy}{dx} = \frac{\sin \alpha}{(1-x \cos \alpha)^2}$
$\frac{dy}{dx} = \frac{\sin \alpha}{x^2 - 2x \cos \alpha + 1}$.
Comparing this with $\frac{dy}{dx} = \frac{m}{x^2+2nx+1}$,we get $m = \sin \alpha$ and $n = -\cos \alpha$.
Therefore,$m^2 + n^2 = \sin^2 \alpha + (-\cos \alpha)^2 = \sin^2 \alpha + \cos^2 \alpha = 1$.

(A) Given the expression $y=\sqrt{(x-\sin x)+\sqrt{(x-\sin x)+\sqrt{(x-\sin x)+\ldots}}}$.
We can rewrite this as $y=\sqrt{(x-\sin x)+y}$.
Squaring both sides,we get $y^2 = x - \sin x + y$.
Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx} = 1 - \cos x + \frac{dy}{dx}$.
Rearranging the terms to solve for $\frac{dy}{dx}$,we have $2y \frac{dy}{dx} - \frac{dy}{dx} = 1 - \cos x$.
Factoring out $\frac{dy}{dx}$,we get $\frac{dy}{dx}(2y-1) = 1 - \cos x$.
Therefore,$\frac{dy}{dx} = \frac{1-\cos x}{2y-1}$.

(C) Given $f(x) = |\log 2 - \sin x|$. Since $\log 2 \approx 0.693$ and $\sin x$ is small near $x=0$,$\log 2 - \sin x > 0$ for $x$ near $0$.
Thus,$f(x) = \log 2 - \sin x$ for $x$ in a neighborhood of $0$.
Then $g(x) = f(f(x)) = \log 2 - \sin(\log 2 - \sin x)$.
Since $g(x)$ is a composition of differentiable functions near $x=0$,it is differentiable at $x=0$.
Now,$g^{\prime}(x) = -\cos(\log 2 - \sin x) \cdot (-\cos x) = \cos(\log 2 - \sin x) \cdot \cos x$.
Evaluating at $x=0$: $g^{\prime}(0) = \cos(\log 2 - \sin 0) \cdot \cos 0 = \cos(\log 2) \cdot 1 = \cos(\log 2)$.

(A) Given equation is $x^{3}+y^{3}-3 a x y=0$.
On differentiating both sides with respect to $x$,we get:
$\frac{d}{d x}(x^{3}) + \frac{d}{d x}(y^{3}) - 3 a \frac{d}{d x}(x y) = 0$
$3 x^{2} + 3 y^{2} \frac{d y}{d x} - 3 a \left( x \frac{d y}{d x} + y \right) = 0$
Dividing by $3$,we get:
$x^{2} + y^{2} \frac{d y}{d x} - a x \frac{d y}{d x} - a y = 0$
Grouping the terms containing $\frac{d y}{d x}$:
$\frac{d y}{d x} (y^{2} - a x) = a y - x^{2}$
Therefore,$\frac{d y}{d x} = \frac{a y - x^{2}}{y^{2} - a x}$.

(D) Given the equation: $x^{\frac{2}{5}} + y^{\frac{2}{5}} = a^{\frac{2}{5}}$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x^{\frac{2}{5}}) + \frac{d}{dx}(y^{\frac{2}{5}}) = \frac{d}{dx}(a^{\frac{2}{5}})$
Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the chain rule:
$\frac{2}{5}x^{\frac{2}{5}-1} + \frac{2}{5}y^{\frac{2}{5}-1} \cdot \frac{dy}{dx} = 0$
$\frac{2}{5}x^{-\frac{3}{5}} + \frac{2}{5}y^{-\frac{3}{5}} \cdot \frac{dy}{dx} = 0$
Dividing by $\frac{2}{5}$:
$x^{-\frac{3}{5}} + y^{-\frac{3}{5}} \cdot \frac{dy}{dx} = 0$
$y^{-\frac{3}{5}} \cdot \frac{dy}{dx} = -x^{-\frac{3}{5}}$
$\frac{dy}{dx} = -\frac{x^{-\frac{3}{5}}}{y^{-\frac{3}{5}}}$
$\frac{dy}{dx} = -\left(\frac{y}{x}\right)^{\frac{3}{5}}$
$\frac{dy}{dx} = -\sqrt[5]{\left(\frac{y}{x}\right)^3}$
Thus,the correct option is $D$.

(B) Given the equation $x^{y} + y^{x} = a^{b}$.
Let $u = x^{y}$ and $v = y^{x}$. Then $u + v = a^{b}$.
Differentiating with respect to $x$,we get $\frac{du}{dx} + \frac{dv}{dx} = 0$.
For $u = x^{y}$,taking log on both sides: $\log u = y \log x$. Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = \frac{y}{x} + \log x \frac{dy}{dx} \implies \frac{du}{dx} = x^{y} (\frac{y}{x} + \log x \frac{dy}{dx})$.
At $x = 1, y = 2$,$u = 1^{2} = 1$,so $\frac{du}{dx} = 1(\frac{2}{1} + \log 1 \frac{dy}{dx}) = 2$.
For $v = y^{x}$,taking log on both sides: $\log v = x \log y$. Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = \log y + \frac{x}{y} \frac{dy}{dx} \implies \frac{dv}{dx} = y^{x} (\log y + \frac{x}{y} \frac{dy}{dx})$.
At $x = 1, y = 2$,$v = 2^{1} = 2$,so $\frac{dv}{dx} = 2(\log 2 + \frac{1}{2} \frac{dy}{dx}) = 2 \log 2 + \frac{dy}{dx}$.
Substituting these into the derivative of the sum: $2 + 2 \log 2 + \frac{dy}{dx} = 0$.
Therefore,$\frac{dy}{dx} = -(2 + 2 \log 2) = -2(1 + \log 2)$.
Wait,re-evaluating the derivative of $v$: $\frac{dv}{dx} = 2(\log 2 + \frac{1}{2} \frac{dy}{dx}) = 2 \log 2 + \frac{dy}{dx}$.
Summing: $2 + 2 \log 2 + \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -2(1 + \log 2)$.
Looking at the options,there seems to be a discrepancy in the denominator. Let's re-check the derivative of $v$ at $x=1, y=2$: $\frac{dv}{dx} = 2(\log 2 + \frac{1}{2} \frac{dy}{dx}) = 2 \log 2 + \frac{dy}{dx}$.
Actually,the derivative of $x^y + y^x = C$ is $\frac{dy}{dx} = -\frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}}$.
At $x=1, y=2$: $\frac{dy}{dx} = -\frac{2(1)^1 + 2^1 \log 2}{1^2 \log 1 + 1(2)^0} = -\frac{2 + 2 \log 2}{0 + 1} = -2(1 + \log 2)$.
Given the options provided,option $B$ is the intended answer if the denominator was meant to be $1$.

(A) Given the equation $y = \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + \dots \infty}}}}$.
Squaring both sides,we get $y^2 = x + \sqrt{y + \sqrt{x + \sqrt{y + \dots \infty}}}$.
Let $u = \sqrt{y + \sqrt{x + \sqrt{y + \dots \infty}}}$. Then $y^2 = x + u$.
Squaring $u$,we get $u^2 = y + \sqrt{x + \sqrt{y + \dots \infty}} = y + y = 2y$.
Thus,$u = \sqrt{2y}$.
Substituting $u$ back into the equation for $y^2$,we have $y^2 = x + \sqrt{2y}$.
Differentiating both sides with respect to $x$:
$2y \frac{dy}{dx} = 1 + \frac{1}{2\sqrt{2y}} \cdot 2 \frac{dy}{dx}$.
$2y \frac{dy}{dx} = 1 + \frac{1}{\sqrt{2y}} \frac{dy}{dx}$.
$\frac{dy}{dx} (2y - \frac{1}{\sqrt{2y}}) = 1$.
$\frac{dy}{dx} (\frac{2y\sqrt{2y} - 1}{\sqrt{2y}}) = 1$.
$\frac{dy}{dx} = \frac{\sqrt{2y}}{2y\sqrt{2y} - 1}$.
Alternatively,from $y^2 - x = \sqrt{2y}$,squaring gives $(y^2 - x)^2 = 2y$.
Differentiating: $2(y^2 - x)(2y \frac{dy}{dx} - 1) = 2 \frac{dy}{dx}$.
$(y^2 - x)(2y \frac{dy}{dx} - 1) = \frac{dy}{dx}$.
$2y(y^2 - x) \frac{dy}{dx} - (y^2 - x) = \frac{dy}{dx}$.
$\frac{dy}{dx} (2y^3 - 2xy - 1) = y^2 - x$.
$\frac{dy}{dx} = \frac{y^2 - x}{2y^3 - 2xy - 1}$.

(A) Given the equation $e^{y} + xy = e$.
At $x = 0$,$e^{y} + 0 = e$,which implies $e^{y} = e$,so $y = 1$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(e^{y} + xy) = \frac{d}{dx}(e)$
$e^{y} \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$.
At $(x, y) = (0, 1)$:
$e^{1} \frac{dy}{dx} + 1 + 0 = 0 \implies e \frac{dy}{dx} = -1 \implies \frac{dy}{dx} = -\frac{1}{e}$.
Differentiating again with respect to $x$:
$\frac{d}{dx}(e^{y} \frac{dy}{dx} + y + x \frac{dy}{dx}) = 0$
$e^{y} (\frac{dy}{dx})^2 + e^{y} \frac{d^2y}{dx^2} + \frac{dy}{dx} + \frac{dy}{dx} + x \frac{d^2y}{dx^2} = 0$
$e^{y} (\frac{dy}{dx})^2 + e^{y} \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} + x \frac{d^2y}{dx^2} = 0$.
Substituting $x = 0, y = 1, \frac{dy}{dx} = -\frac{1}{e}$:
$e(-\frac{1}{e})^2 + e \frac{d^2y}{dx^2} + 2(-\frac{1}{e}) + 0 = 0$
$\frac{1}{e} + e \frac{d^2y}{dx^2} - \frac{2}{e} = 0$
$e \frac{d^2y}{dx^2} = \frac{1}{e} \implies \frac{d^2y}{dx^2} = \frac{1}{e^2}$.
The ordered pair is $(-\frac{1}{e}, \frac{1}{e^2})$.

(B) Given the equation: $x \cdot \log_{e}(\log_{e} x) - x^2 + y^2 = 4$.
First,find the value of $y$ at $x=e$:
$e \cdot \log_{e}(\log_{e} e) - e^2 + y^2 = 4$.
Since $\log_{e} e = 1$ and $\log_{e} 1 = 0$,we have $e \cdot 0 - e^2 + y^2 = 4$,which implies $y^2 = 4 + e^2$.
Since $y > 0$,$y = \sqrt{4 + e^2}$.
Now,differentiate the given equation with respect to $x$:
$\frac{d}{dx}[x \cdot \log_{e}(\log_{e} x)] - \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = 0$.
Using the product rule on the first term:
$1 \cdot \log_{e}(\log_{e} x) + x \cdot \frac{1}{\log_{e} x} \cdot \frac{1}{x} - 2x + 2y \frac{dy}{dx} = 0$.
$\log_{e}(\log_{e} x) + \frac{1}{\log_{e} x} - 2x + 2y \frac{dy}{dx} = 0$.
Substitute $x=e$:
$\log_{e}(\log_{e} e) + \frac{1}{\log_{e} e} - 2e + 2y \frac{dy}{dx} = 0$.
$\log_{e}(1) + \frac{1}{1} - 2e + 2y \frac{dy}{dx} = 0$.
$0 + 1 - 2e + 2y \frac{dy}{dx} = 0$.
$2y \frac{dy}{dx} = 2e - 1$.
$\frac{dy}{dx} = \frac{2e - 1}{2y}$.
Substituting $y = \sqrt{4 + e^2}$,we get $\frac{dy}{dx} = \frac{2e - 1}{2\sqrt{4 + e^2}}$.

(C) Given equation: $\log(x+y)=2xy$ ... $(i)$
At $x=0$,$\log(0+y)=2(0)y$,which implies $\log(y)=0$,so $y=e^0=1$.
Differentiating both sides of $(i)$ with respect to $x$:
$\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) = 2y + 2x \frac{dy}{dx}$
Substitute $x=0$ and $y=1$ into the differentiated equation:
$\frac{1}{0+1} \left(1 + \frac{dy}{dx}\right) = 2(1) + 2(0) \frac{dy}{dx}$
$1 + \frac{dy}{dx} = 2$
$\frac{dy}{dx} = 2 - 1 = 1$
Thus,the value of $\frac{dy}{dx}$ at $x=0$ is $1$.

(B) Given the equation: $(2x)^{2y} = 4e^{2x-2y}$.
Taking the natural logarithm $(\log_e)$ on both sides:
$2y \log_e(2x) = \log_e 4 + (2x - 2y) \log_e e$
Since $\log_e 4 = 2 \log_e 2$ and $\log_e e = 1$,we have:
$2y \log_e(2x) = 2 \log_e 2 + 2x - 2y$
Dividing by $2$:
$y \log_e(2x) = \log_e 2 + x - y$
Rearranging to solve for $y$:
$y(1 + \log_e 2x) = x + \log_e 2$
$y = \frac{x + \log_e 2}{1 + \log_e 2x} \quad \dots(i)$
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} (1 + \log_e 2x) + y \left( \frac{1}{2x} \cdot 2 \right) = 1 - \frac{dy}{dx}$
$\frac{dy}{dx} (1 + \log_e 2x) + \frac{y}{x} = 1 - \frac{dy}{dx}$
$\frac{dy}{dx} (1 + \log_e 2x + 1) = 1 - \frac{y}{x}$
$\frac{dy}{dx} (2 + \log_e 2x) = \frac{x - y}{x}$
Wait,let us re-evaluate the derivative step:
From $y(1 + \log_e 2x) = x + \log_e 2$,differentiate:
$\frac{dy}{dx} (1 + \log_e 2x) + y \left( \frac{1}{x} \right) = 1$
$\frac{dy}{dx} (1 + \log_e 2x) = 1 - \frac{y}{x} = \frac{x - y}{x}$
Substitute $y$ from $(i)$:
$\frac{dy}{dx} (1 + \log_e 2x) = \frac{x - \frac{x + \log_e 2}{1 + \log_e 2x}}{x} = \frac{x(1 + \log_e 2x) - (x + \log_e 2)}{x(1 + \log_e 2x)}$
$= \frac{x + x \log_e 2x - x - \log_e 2}{x(1 + \log_e 2x)} = \frac{x \log_e 2x - \log_e 2}{x(1 + \log_e 2x)}$
Multiplying both sides by $(1 + \log_e 2x)$:
$(1 + \log_e 2x)^2 \frac{dy}{dx} = \frac{x \log_e 2x - \log_e 2}{x}$

(B) Given $y^{\frac{1}{m}}+y^{-\frac{1}{m}}=2x$.
Squaring both sides,we get $(y^{\frac{1}{m}}+y^{-\frac{1}{m}})^2 = (2x)^2$,which implies $y^{\frac{2}{m}}+y^{-\frac{2}{m}}+2 = 4x^2$,so $y^{\frac{2}{m}}+y^{-\frac{2}{m}} = 4x^2-2$ ... $(1)$
Differentiating both sides with respect to $y$,we get $\frac{1}{m} y^{\frac{1}{m}-1} - \frac{1}{m} y^{-\frac{1}{m}-1} = 2 \frac{dx}{dy}$.
Multiplying by $my$,we get $y^{\frac{1}{m}} - y^{-\frac{1}{m}} = 2my \frac{dx}{dy} = \frac{2my}{dy/dx}$.
Squaring both sides,we get $(y^{\frac{1}{m}}-y^{-\frac{1}{m}})^2 = \frac{4m^2y^2}{(dy/dx)^2}$.
This simplifies to $y^{\frac{2}{m}}+y^{-\frac{2}{m}}-2 = \frac{4m^2y^2}{(dy/dx)^2}$ ... $(2)$
Subtracting $(2)$ from $(1)$,we get $(y^{\frac{2}{m}}+y^{-\frac{2}{m}}+2) - (y^{\frac{2}{m}}+y^{-\frac{2}{m}}-2) = 4x^2 - \frac{4m^2y^2}{(dy/dx)^2}$.
$4 = 4x^2 - \frac{4m^2y^2}{(dy/dx)^2}$.
Dividing by $4$,we get $1 = x^2 - \frac{m^2y^2}{(dy/dx)^2}$.
Rearranging the terms,we get $\frac{m^2y^2}{(dy/dx)^2} = x^2-1$.
Therefore,$(x^2-1)(\frac{dy}{dx})^2 = m^2y^2$.

(A) Given the equation: $e^x + e^y = e^{x+y}$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(e^x) + \frac{d}{dx}(e^y) = \frac{d}{dx}(e^{x+y})$
$e^x + e^y \cdot \frac{dy}{dx} = e^{x+y} \cdot (1 + \frac{dy}{dx})$
$e^x + e^y \cdot \frac{dy}{dx} = e^{x+y} + e^{x+y} \cdot \frac{dy}{dx}$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$e^y \cdot \frac{dy}{dx} - e^{x+y} \cdot \frac{dy}{dx} = e^{x+y} - e^x$
$\frac{dy}{dx} (e^y - e^{x+y}) = e^{x+y} - e^x$
Since $e^{x+y} = e^x + e^y$,substitute this into the equation:
$\frac{dy}{dx} (e^y - (e^x + e^y)) = (e^x + e^y) - e^x$
$\frac{dy}{dx} (e^y - e^x - e^y) = e^y$
$\frac{dy}{dx} (-e^x) = e^y$
$\frac{dy}{dx} = -\frac{e^y}{e^x} = -e^{y-x}$

(A) We know the trigonometric identity $\tan^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for all $u \in \mathbb{R}$.
Given the equation $xy = \tan^{-1}(xy) + \cot^{-1}(xy)$,we can substitute the identity to get $xy = \frac{\pi}{2}$.
Differentiating both sides with respect to $x$ using the product rule:
$y + x \frac{dy}{dx} = 0$.
Solving for $\frac{dy}{dx}$,we get $\frac{dy}{dx} = -\frac{y}{x}$.
Evaluating this at the point $(4, 2)$:
$\left(\frac{dy}{dx}\right)_{(4,2)} = -\frac{2}{4} = -\frac{1}{2}$.

(C) Given the equation: $\log (x+y) = \log (xy) + 3$
Using the property of logarithms $\log a - \log b = \log (\frac{a}{b})$,we get:
$\log (x+y) - \log (xy) = 3$
$\log \left(\frac{x+y}{xy}\right) = 3$
Converting from logarithmic to exponential form:
$\frac{x+y}{xy} = e^3$
$\frac{x}{xy} + \frac{y}{xy} = e^3$
$\frac{1}{y} + \frac{1}{x} = e^3$
Differentiating both sides with respect to $x$:
$\frac{d}{dx} (y^{-1}) + \frac{d}{dx} (x^{-1}) = \frac{d}{dx} (e^3)$
$-y^{-2} \frac{dy}{dx} - x^{-2} = 0$
$-\frac{1}{y^2} \frac{dy}{dx} = \frac{1}{x^2}$
$\frac{dy}{dx} = -\frac{y^2}{x^2} = -\left(\frac{y}{x}\right)^2$

(A) Given the equation: $\log (x+y) = \log (xy) + a$
Using logarithmic properties,we can rewrite this as: $\log (x+y) - \log (xy) = a$
$\log \left( \frac{x+y}{xy} \right) = a$
$\frac{x+y}{xy} = e^a$
$\frac{1}{y} + \frac{1}{x} = e^a$
Differentiating both sides with respect to $x$:
$-\frac{1}{y^2} \frac{dy}{dx} - \frac{1}{x^2} = 0$
$\frac{dy}{dx} = -\frac{y^2}{x^2}$
Substituting $x=2$ and $y=4$:
$\frac{dy}{dx} = -\frac{4^2}{2^2} = -\frac{16}{4} = -4$

(D) Given the equation $e^{-y} \cdot y = x$.
Taking the derivative with respect to $x$ on both sides:
$\frac{d}{dx}(y e^{-y}) = \frac{d}{dx}(x)$
Using the product rule:
$y \cdot \frac{d}{dx}(e^{-y}) + e^{-y} \cdot \frac{dy}{dx} = 1$
$y \cdot (-e^{-y}) \frac{dy}{dx} + e^{-y} \frac{dy}{dx} = 1$
Factor out $\frac{dy}{dx}$:
$\frac{dy}{dx} (e^{-y} - y e^{-y}) = 1$
$\frac{dy}{dx} e^{-y} (1 - y) = 1$
$\frac{dy}{dx} = \frac{1}{e^{-y}(1 - y)}$
Since $e^{-y} = \frac{x}{y}$,substitute this into the expression:
$\frac{dy}{dx} = \frac{1}{(\frac{x}{y})(1 - y)} = \frac{y}{x(1 - y)}$.

(B) Given equation: $y = x \tan y$
Differentiating both sides with respect to $x$:
$\frac{d y}{d x} = x \sec^2 y \frac{d y}{d x} + \tan y$
Rearranging the terms to isolate $\frac{d y}{d x}$:
$\frac{d y}{d x} - x \sec^2 y \frac{d y}{d x} = \tan y$
$\frac{d y}{d x} (1 - x \sec^2 y) = \tan y$
$\frac{d y}{d x} = \frac{\tan y}{1 - x \sec^2 y}$
Multiply numerator and denominator by $x$:
$\frac{d y}{d x} = \frac{x \tan y}{x - x^2 \sec^2 y}$
Since $y = x \tan y$,we have $\tan y = \frac{y}{x}$:
$\frac{d y}{d x} = \frac{y}{x - x^2 (1 + \tan^2 y)} = \frac{y}{x - x^2 - x^2 \tan^2 y}$
Substituting $x^2 \tan^2 y = y^2$:
$\frac{d y}{d x} = \frac{y}{x - x^2 - y^2}$

(A) Given the equation: $y = 1 + xe^y$
Differentiating both sides with respect to $x$ using the product rule:
$\frac{dy}{dx} = 0 + \left( x \cdot e^y \frac{dy}{dx} + e^y \cdot 1 \right)$
$\frac{dy}{dx} = xe^y \frac{dy}{dx} + e^y$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx} - xe^y \frac{dy}{dx} = e^y$
$\frac{dy}{dx} (1 - xe^y) = e^y$
$\frac{dy}{dx} = \frac{e^y}{1 - xe^y}$
From the original equation,we know $xe^y = y - 1$. Substituting this into the expression:
$\frac{dy}{dx} = \frac{e^y}{1 - (y - 1)}$
$\frac{dy}{dx} = \frac{e^y}{1 - y + 1}$
$\frac{dy}{dx} = \frac{e^y}{2 - y}$

(C) Given equation: $\sin ^2 x + \cos ^2 y = 1$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\sin ^2 x) + \frac{d}{dx}(\cos ^2 y) = \frac{d}{dx}(1)$
$2 \sin x \cos x + 2 \cos y (-\sin y) \frac{dy}{dx} = 0$
$\sin 2x - \sin 2y \frac{dy}{dx} = 0$
$\sin 2y \frac{dy}{dx} = \sin 2x$
$\frac{dy}{dx} = \frac{\sin 2x}{\sin 2y}$

(A) Given the equation $\sqrt{x} + \sqrt{y} = \sqrt{xy}$.
Dividing both sides by $\sqrt{xy}$,we get:
$\frac{\sqrt{x}}{\sqrt{xy}} + \frac{\sqrt{y}}{\sqrt{xy}} = 1$
$\frac{1}{\sqrt{y}} + \frac{1}{\sqrt{x}} = 1$
$y^{-1/2} + x^{-1/2} = 1$
Differentiating both sides with respect to $x$:
$-\frac{1}{2} y^{-3/2} \frac{dy}{dx} - \frac{1}{2} x^{-3/2} = 0$
$-\frac{1}{2 y^{3/2}} \frac{dy}{dx} = \frac{1}{2 x^{3/2}}$
$\frac{dy}{dx} = -\frac{2 y^{3/2}}{2 x^{3/2}}$
$\frac{dy}{dx} = -\left(\frac{y}{x}\right)^{3/2}$

(B) Given $\sin \left(\frac{x+y}{x-y}\right)=\tan \frac{\pi}{5}$.
Let $K = \sin^{-1}(\tan \frac{\pi}{5})$,which is a constant.
Then $\frac{x+y}{x-y} = K$.
$x+y = K(x-y)$.
Differentiating both sides with respect to $x$:
$1 + \frac{dy}{dx} = K(1 - \frac{dy}{dx})$.
$1 + \frac{dy}{dx} = K - K\frac{dy}{dx}$.
$\frac{dy}{dx}(1+K) = K-1$.
$\frac{dy}{dx} = \frac{K-1}{K+1}$.
Substituting $K = \frac{x+y}{x-y}$ back into the expression:
$\frac{dy}{dx} = \frac{\frac{x+y}{x-y} - 1}{\frac{x+y}{x-y} + 1} = \frac{x+y - (x-y)}{x+y + (x-y)} = \frac{2y}{2x} = \frac{y}{x}$.

(A) Given the equation $y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$.
Substitute $x = \sin \alpha$ and $y = \sin \beta$,where $\alpha = \sin^{-1} x$ and $\beta = \sin^{-1} y$.
The equation becomes $\sin \beta \cos \alpha + \sin \alpha \cos \beta = 1$.
Using the trigonometric identity $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$,we get $\sin(\alpha + \beta) = 1$.
Thus,$\alpha + \beta = \sin^{-1}(1) = \frac{\pi}{2}$.
Substituting back,we have $\sin^{-1} x + \sin^{-1} y = \frac{\pi}{2}$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\sin^{-1} x) + \frac{d}{dx}(\sin^{-1} y) = \frac{d}{dx}(\frac{\pi}{2})$.
$\frac{1}{\sqrt{1-x^{2}}} + \frac{1}{\sqrt{1-y^{2}}} \frac{dy}{dx} = 0$.
Rearranging the terms:
$\frac{1}{\sqrt{1-y^{2}}} \frac{dy}{dx} = -\frac{1}{\sqrt{1-x^{2}}}$.
Therefore,$\frac{dy}{dx} = -\sqrt{\frac{1-y^{2}}{1-x^{2}}}$.

(A) Given the equation: $\frac{x}{\sqrt{1+x}} + \frac{y}{\sqrt{1+y}} = 0$
Rearranging the terms: $\frac{x}{\sqrt{1+x}} = -\frac{y}{\sqrt{1+y}}$
Squaring both sides: $\frac{x^2}{1+x} = \frac{y^2}{1+y}$
Cross-multiplying: $x^2(1+y) = y^2(1+x)$
$x^2 + x^2y = y^2 + xy^2$
$x^2 - y^2 + x^2y - xy^2 = 0$
$(x-y)(x+y) + xy(x-y) = 0$
Since $x \neq y$,we can divide by $(x-y)$:
$x + y + xy = 0$
$y(1+x) = -x$
$y = -\frac{x}{1+x}$
Now,differentiate with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = -\frac{(1+x)(1) - x(1)}{(1+x)^2}$
$\frac{dy}{dx} = -\frac{1+x-x}{(1+x)^2}$
$\frac{dy}{dx} = -\frac{1}{(1+x)^2}$

(B) Given: $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4$
$\therefore \frac{x+y}{\sqrt{x y}}=4 \Rightarrow x+y=4 \sqrt{x y}$
Squaring both sides,we get:
$(x+y)^{2}=16 x y \Rightarrow x^{2}+2 x y+y^{2}=16 x y \Rightarrow x^{2}+y^{2}=14 x y$
Differentiating both sides with respect to $x$:
$2 x+2 y \frac{d y}{d x}=14 \left(x \frac{d y}{d x}+y\right)$
Dividing by $2$:
$x+y \frac{d y}{d x}=7 x \frac{d y}{d x}+7 y$
Rearranging the terms to solve for $\frac{d y}{d x}$:
$y \frac{d y}{d x}-7 x \frac{d y}{d x}=7 y-x$
$(y-7 x) \frac{d y}{d x}=7 y-x$
$\frac{d y}{d x}=\frac{7 y-x}{y-7 x}$

(C) Given equation is $x^y = e^{x - y}$.
Taking $\log$ on both sides,we get $y \log x = (x - y) \log e = x - y$ . . . . . . $(i)$.
When $x = 1$,substituting in $(i)$ gives $y \log 1 = 1 - y$,which implies $0 = 1 - y$,so $y = 1$.
Differentiating both sides of $(i)$ with respect to $x$ using the product rule:
$y \cdot (\frac{1}{x}) + \log x \cdot \frac{dy}{dx} = 1 - \frac{dy}{dx}$.
Rearranging the terms to solve for $\frac{dy}{dx}$:
$\frac{dy}{dx} (\log x + 1) = 1 - \frac{y}{x}$.
$\frac{dy}{dx} = \frac{x - y}{x (\log x + 1)}$.
At $x = 1$ and $y = 1$:
$\frac{dy}{dx} = \frac{1 - 1}{1 (\log 1 + 1)} = \frac{0}{1} = 0$.

(D) Given the equation: $x^{p} + y^{q} = (x + y)^{p + q}$.
Taking the natural logarithm on both sides,we get: $p \ln x + q \ln y = (p + q) \ln (x + y)$.
Differentiating both sides with respect to $x$:
$\frac{p}{x} + \frac{q}{y} \cdot \frac{dy}{dx} = \frac{p + q}{x + y} \left( 1 + \frac{dy}{dx} \right)$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{p}{x} - \frac{p + q}{x + y} = \left( \frac{p + q}{x + y} - \frac{q}{y} \right) \frac{dy}{dx}$.
Simplifying both sides:
$\frac{p(x + y) - x(p + q)}{x(x + y)} = \left( \frac{y(p + q) - q(x + y)}{y(x + y)} \right) \frac{dy}{dx}$.
$\frac{px + py - px - qx}{x(x + y)} = \left( \frac{py + qy - qx - qy}{y(x + y)} \right) \frac{dy}{dx}$.
$\frac{py - qx}{x(x + y)} = \frac{py - qx}{y(x + y)} \cdot \frac{dy}{dx}$.
Canceling the common term $(py - qx)$ and $(x + y)$ from both sides,we get:
$\frac{1}{x} = \frac{1}{y} \cdot \frac{dy}{dx}$.
Therefore,$\frac{dy}{dx} = \frac{y}{x}$.

(B) Given: $x^2+y^2=t+\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$.
Squaring the first equation on both sides,we get:
$(x^2+y^2)^2 = (t+\frac{1}{t})^2$
$x^4+y^4+2x^2y^2 = t^2+\frac{1}{t^2}+2$.
Since $x^4+y^4=t^2+\frac{1}{t^2}$,we substitute this into the equation:
$(t^2+\frac{1}{t^2})+2x^2y^2 = t^2+\frac{1}{t^2}+2$.
This simplifies to $2x^2y^2 = 2$,which implies $x^2y^2 = 1$.
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 equations are:
$x^{2}+y^{2}=t+\frac{1}{t}$ ... $(1)$
$x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}}$ ... $(2)$
Squaring equation $(1)$ on both sides,we get:
$(x^{2}+y^{2})^{2} = (t+\frac{1}{t})^{2}$
$x^{4}+y^{4}+2x^{2}y^{2} = t^{2}+\frac{1}{t^{2}}+2$
Substituting the value from equation $(2)$ into this result:
$(t^{2}+\frac{1}{t^{2}}) + 2x^{2}y^{2} = t^{2}+\frac{1}{t^{2}}+2$
$2x^{2}y^{2} = 2$
$x^{2}y^{2} = 1$
Differentiating both sides with respect to $x$ using the product rule:
$x^{2}(2y \frac{dy}{dx}) + y^{2}(2x) = 0$
$2x^{2}y \frac{dy}{dx} = -2xy^{2}$
$\frac{dy}{dx} = \frac{-2xy^{2}}{2x^{2}y} = -\frac{y}{x}$

(D) Given equation: $\log (x+y) = \sin (x+y)$
Differentiating both sides with respect to $x$:
$\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) = \cos (x+y) \left(1 + \frac{dy}{dx}\right)$
Rearranging the terms:
$\left(1 + \frac{dy}{dx}\right) \left(\frac{1}{x+y} - \cos (x+y)\right) = 0$
Since $\log (x+y) = \sin (x+y)$,the term $\frac{1}{x+y} - \cos (x+y)$ cannot be zero for all $x, y$ in the domain.
Therefore,we must have:
$1 + \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -1$

(A) Given the equation: $\log (x+y)=2xy \quad ...(i)$
Differentiating both sides with respect to $x$ using the chain rule and product rule:
$\frac{1}{x+y} \cdot (1 + y^{\prime}) = 2(x y^{\prime} + y)$
Rearranging to solve for $y^{\prime}$:
$1 + y^{\prime} = 2(x+y)(x y^{\prime} + y)$
$1 + y^{\prime} = 2x^2 y^{\prime} + 2xy + 2xy y^{\prime} + 2y^2$
$y^{\prime}(1 - 2x^2 - 2xy) = 2xy + 2y^2 - 1$
$y^{\prime} = \frac{2xy + 2y^2 - 1}{1 - 2x^2 - 2xy}$
Now,find the value of $y$ when $x=0$ from equation $(i)$:
$\log(0+y) = 2(0)y$
$\log(y) = 0$
$y = e^0 = 1$
Substitute $x=0$ and $y=1$ into the expression for $y^{\prime}$:
$y^{\prime}(0) = \frac{2(0)(1) + 2(1)^2 - 1}{1 - 2(0)^2 - 2(0)(1)}$
$y^{\prime}(0) = \frac{0 + 2 - 1}{1 - 0 - 0} = \frac{1}{1} = 1$

(C) Given $x = e^{(x/y)}$.
Taking natural logarithm on both sides,we get $\log x = \frac{x}{y}$.
This implies $y \log x = x$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{dy}{dx} \cdot \log x + y \cdot \frac{1}{x} = 1$.
Multiplying the entire equation by $x$:
$x \log x \cdot \frac{dy}{dx} + y = x$.
Rearranging the terms to solve for $\frac{dy}{dx}$:
$x \log x \cdot \frac{dy}{dx} = x - y$.
Therefore,$\frac{dy}{dx} = \frac{x-y}{x \log x}$.

(B) Given: $(2x)^{2y} = 4e^{2x-2y}$
Taking natural logarithm on both sides:
$2y \log(2x) = \log 4 + 2x - 2y$
$2y \log(2x) = 2 \log 2 + 2x - 2y$
Dividing by $2$:
$y \log(2x) = \log 2 + x - y$
$y(1 + \log 2x) = x + \log 2$
$y = \frac{x + \log 2}{1 + \log 2x}$ ... $(1)$
Now,differentiating $y \log(2x) = \log 2 + x - y$ with respect to $x$:
$\frac{dy}{dx} \log(2x) + y \cdot \frac{1}{2x} \cdot 2 = 0 + 1 - \frac{dy}{dx}$
$\frac{dy}{dx} \log(2x) + \frac{y}{x} = 1 - \frac{dy}{dx}$
$\frac{dy}{dx} (1 + \log 2x) = 1 - \frac{y}{x} = \frac{x - y}{x}$
Substitute $y$ from $(1)$:
$\frac{dy}{dx} (1 + \log 2x) = \frac{x - \frac{x + \log 2}{1 + \log 2x}}{x} = \frac{x(1 + \log 2x) - x - \log 2}{x(1 + \log 2x)}$
$\frac{dy}{dx} (1 + \log 2x) = \frac{x + x \log 2x - x - \log 2}{x(1 + \log 2x)} = \frac{x \log 2x - \log 2}{x(1 + \log 2x)}$
Multiplying both sides by $(1 + \log 2x)$:
$(1 + \log 2x)^2 \frac{dy}{dx} = \frac{x \log 2x - \log 2}{x}$

(D) Given the equation: $(2x)^{2y} = 4e^{2x-2y}$
Taking the natural logarithm on both sides:
$2y \log_e(2x) = \log_e(4) + 2x - 2y$
$2y \log_e(2x) + 2y = 2x + \log_e(4)$
$2y(1 + \log_e(2x)) = 2x + 2 \log_e(2)$
$y(1 + \log_e(2x)) = x + \log_e(2)$
$y = \frac{x + \log_e(2)}{1 + \log_e(2x)}$
Differentiating both sides with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = \frac{(1 + \log_e(2x)) \cdot \frac{d}{dx}(x + \log_e(2)) - (x + \log_e(2)) \cdot \frac{d}{dx}(1 + \log_e(2x))}{(1 + \log_e(2x))^2}$
$\frac{dy}{dx} = \frac{(1 + \log_e(2x)) \cdot 1 - (x + \log_e(2)) \cdot \frac{1}{2x} \cdot 2}{(1 + \log_e(2x))^2}$
$\frac{dy}{dx} = \frac{1 + \log_e(2x) - \frac{x + \log_e(2)}{x}}{(1 + \log_e(2x))^2}$
$(1 + \log_e(2x))^2 \frac{dy}{dx} = 1 + \log_e(2x) - 1 - \frac{\log_e(2)}{x}$
$(1 + \log_e(2x))^2 \frac{dy}{dx} = \log_e(2x) - \frac{\log_e(2)}{x} = \frac{x \log_e(2x) - \log_e(2)}{x}$

(A) Given the equation $x^y \cdot y^x = 16$.
Taking the natural logarithm on both sides,we get:
$y \ln x + x \ln y = \ln 16$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{d}{dx}(y \ln x) + \frac{d}{dx}(x \ln y) = \frac{d}{dx}(\ln 16)$.
$\left( \frac{dy}{dx} \ln x + y \cdot \frac{1}{x} \right) + \left( 1 \cdot \ln y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} \right) = 0$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx} (\ln x + \frac{x}{y}) = -(\frac{y}{x} + \ln y)$.
$\frac{dy}{dx} = -\frac{\frac{y}{x} + \ln y}{\ln x + \frac{x}{y}}$.
Now,substitute the point $(2, 2)$ into the expression:
$\left( \frac{dy}{dx} \right)_{(2, 2)} = -\frac{\frac{2}{2} + \ln 2}{\ln 2 + \frac{2}{2}} = -\frac{1 + \ln 2}{1 + \ln 2} = -1$.

(C) Given $\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2$.
By the definition of logarithm,$\frac{x^{3}-y^{3}}{x^{3}+y^{3}}=10^{2}=100$.
Therefore,$x^{3}-y^{3}=100(x^{3}+y^{3})$.
$x^{3}-y^{3}=100x^{3}+100y^{3}$.
$-99x^{3}=101y^{3}$.
Differentiating both sides with respect to $y$:
$-99 \cdot 3x^{2} \frac{dx}{dy} = 101 \cdot 3y^{2}$.
$-297x^{2} \frac{dx}{dy} = 303y^{2}$.
$\frac{dx}{dy} = -\frac{303y^{2}}{297x^{2}} = -\frac{101y^{2}}{99x^{2}}$.
Thus,$\frac{dx}{dy} = \left(-\frac{101}{99}\right) \frac{y^{2}}{x^{2}}$.

(D) Given,$\log _{10}\left(\frac{x^3-y^3}{x^3+y^3}\right)=2$
$\Rightarrow \frac{x^3-y^3}{x^3+y^3} = 10^2 = 100$
$\Rightarrow x^3 - y^3 = 100x^3 + 100y^3$
$\Rightarrow -99x^3 = 101y^3$
$\Rightarrow y^3 = -\frac{99}{101}x^3$
Differentiating both sides with respect to $x$:
$3y^2 \frac{dy}{dx} = -\frac{99}{101} \cdot 3x^2$
$\frac{dy}{dx} = -\frac{99}{101} \cdot \frac{x^2}{y^2}$
Since $y^3 = -\frac{99}{101}x^3$,we have $-\frac{99}{101} = \frac{y^3}{x^3}$
Substituting this into the derivative:
$\frac{dy}{dx} = \frac{y^3}{x^3} \cdot \frac{x^2}{y^2} = \frac{y}{x}$

(A) Given,$\log _{10}\left(\frac{x^2-y^2}{x^2+y^2}\right)=2$.
Applying the definition of logarithm,we get $\frac{x^2-y^2}{x^2+y^2} = 10^2 = 100$.
Cross-multiplying,we have $x^2 - y^2 = 100(x^2 + y^2)$.
$x^2 - y^2 = 100x^2 + 100y^2$.
Rearranging the terms,we get $x^2 - 100x^2 = 100y^2 + y^2$,which simplifies to $-99x^2 = 101y^2$.
Now,differentiate both sides with respect to $x$:
$\frac{d}{dx}(-99x^2) = \frac{d}{dx}(101y^2)$.
$-99(2x) = 101(2y) \frac{dy}{dx}$.
$-198x = 202y \frac{dy}{dx}$.
Therefore,$\frac{dy}{dx} = -\frac{198x}{202y} = -\frac{99x}{101y}$.

(C) Given equation: $e^{x} + e^{y} = e^{x+y}$.
Divide both sides by $e^{x+y}$:
$\frac{e^{x}}{e^{x+y}} + \frac{e^{y}}{e^{x+y}} = 1$
$e^{-y} + e^{-x} = 1$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(e^{-y}) + \frac{d}{dx}(e^{-x}) = \frac{d}{dx}(1)$
$-e^{-y} \frac{dy}{dx} - e^{-x} = 0$
$-e^{-y} \frac{dy}{dx} = e^{-x}$
$\frac{dy}{dx} = -\frac{e^{-x}}{e^{-y}}$
$\frac{dy}{dx} = -e^{y-x}$

(D) We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.
Given equation is $x^2 y^2 = \frac{\pi}{2}$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(\frac{\pi}{2})$
$2x y^2 + x^2 (2y \frac{dy}{dx}) = 0$
Divide by $2x$ (assuming $x \neq 0$):
$y^2 + xy \frac{dy}{dx} = 0$
Substitute $x = 1$ and $y = 2$:
$(2)^2 + (1)(2) \frac{dy}{dx} = 0$
$4 + 2 \frac{dy}{dx} = 0$
$2 \frac{dy}{dx} = -4$
$\frac{dy}{dx} = -2$

(D) Given: $x^2+y^2=t+\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$.
Squaring the first equation: $(x^2+y^2)^2 = (t+\frac{1}{t})^2$.
$x^4+y^4+2x^2y^2 = t^2+\frac{1}{t^2}+2$.
Substituting $x^4+y^4 = t^2+\frac{1}{t^2}$ into the equation:
$(t^2+\frac{1}{t^2}) + 2x^2y^2 = t^2+\frac{1}{t^2}+2$.
$2x^2y^2 = 2 \implies x^2y^2 = 1$.
Thus,$y^2 = \frac{1}{x^2}$.
Differentiating both sides with respect to $x$:
$2y \frac{dy}{dx} = -2x^{-3} = -\frac{2}{x^3}$.
$\frac{dy}{dx} = -\frac{1}{x^3y}$.

(C) Given the equation $x^{2}+y^{2}=1$.
Differentiating both sides with respect to $y$,we get:
$2x \frac{dx}{dy} + 2y = 0$
$2x \frac{dx}{dy} = -2y$
$\frac{dx}{dy} = -\frac{y}{x}$
Now,differentiating $\frac{dx}{dy}$ with respect to $y$ using the quotient rule:
$\frac{d^{2}x}{dy^{2}} = \frac{d}{dy} \left( -\frac{y}{x} \right) = -\left[ \frac{x(1) - y(\frac{dx}{dy})}{x^{2}} \right]$
Substitute $\frac{dx}{dy} = -\frac{y}{x}$ into the expression:
$\frac{d^{2}x}{dy^{2}} = -\left[ \frac{x - y(-\frac{y}{x})}{x^{2}} \right]$
$\frac{d^{2}x}{dy^{2}} = -\left[ \frac{x + \frac{y^{2}}{x}}{x^{2}} \right] = -\left[ \frac{x^{2} + y^{2}}{x^{3}} \right]$
Since $x^{2}+y^{2}=1$,we have:
$\frac{d^{2}x}{dy^{2}} = -\frac{1}{x^{3}}$