⇒Derivative

Correct Option is (A)

$\log (x+y)=2xy$ .... (i)

Differentiating w.r.t. $x$, we get

$\begin{array}{ll}& \frac{1}{x+y}(1+\frac{\mathrm{d}y}{\mathrm{d}x})=2y+2x\frac{\mathrm{d}y}{\mathrm{d}x}\\ \therefore & \frac{1}{x+y}+\frac{1}{x+y}\frac{\mathrm{d}y}{\mathrm{d}x}=2y+2x\frac{\mathrm{d}y}{\mathrm{d}x}\text{.... (ii)}\\ & \text{ At }x=0,\text{ (i) }\Rightarrow y=1\\ \therefore & \text{ (ii) }{\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}|}_{x=0}=1\end{array}$

Correct Option is (C)

$\log (x+y)=2xy$ ...... (i)

Differentiating both sides w.r.t. $x$, we get

$\begin{array}{rl}& \left(\frac{1}{x+y}\right)(1+\frac{\mathrm{d}y}{\mathrm{d}x})=2(x\frac{\mathrm{d}y}{\mathrm{d}x}+y)\\ & \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1-2xy-2y^2}{2x^2+2xy-1}\\ & \text{ Putting }x=0\text{ in (i), we get }\\ & y=1\\ \therefore & {\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}_{x=0}=\frac{1-0-2}{0+0-1}=1\end{array}$

Correct Option is (A)

$\log (x+y)=2xy$

Differentiating w.r.t. $x$, we get

$\begin{array}{rl}& \frac{1}{x+y}(1+\frac{\mathrm{d}y}{\mathrm{d}x})=2x\frac{\mathrm{d}y}{\mathrm{d}x}+2y\\ & \frac{1}{x+y}+\frac{1}{(x+y)}\frac{\mathrm{d}y}{\mathrm{d}x}=2x\frac{\mathrm{d}y}{\mathrm{d}x}+2y\\ & (\frac{1}{x+y}-2x)\frac{\mathrm{d}y}{\mathrm{d}x}=2y-\frac{1}{x+y}\\ & \frac{\mathrm{d}y}{\mathrm{d}x}(\frac{1}{x+y}-2x)=2y-\frac{1}{x+y}\\ & \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(2y-\frac{1}{x+y})}{(\frac{1}{x+y}-2x)}\\ & \text{ For }x=0,\log (y)=0\\ & \Rightarrow y=1\\ & {\frac{\mathrm{d}y}{\mathrm{d}x}|}_{(0,1)}=\frac{(2-\frac{1}{0+1})}{(\frac{1}{0+1}-0)}=1\end{array}$

Correct Option is (B)

$x^2+y^2=t+\frac{1}{t}$

Squaring on both sides, we get

$x^4+y^4+2x^2{y}^2=t^2+\frac{1}{t^2}+2$

$\begin{array}{l}\therefore {\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}+2x^2{y}^2={\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}+2...[\because x^4+y^4=t^2+\frac{1}{t^2},\text{given}]\\ \\ 2x^2{y}^2=2\\ x^2{y}^2=1\\ \text{ differentiating w.r.t. }x,\text{ we get }\\ x^{2}2y\frac{\mathrm{d}y}{\mathrm{d}x}+2x{y}^2=0\\ x^{2}2y\frac{\mathrm{d}y}{\mathrm{d}x}=-2x{y}^2\\ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{-2x{y}^2}{2x^{2}y}=\frac{-y}{x}\end{array}$

Correct Option is (D)

$\begin{array}{rl}& e^{-y}\cdot y=x\\ & \therefore \frac{y}{e^y}=x\Rightarrow y=x{e}^y\text{....(1) and }{e}^y=\frac{y}{x}\text{....(2)}\\ & \text{ Now }y=x{e}^y\\ & \therefore \frac{dy}{dx}=x{e}^y\frac{dy}{dx}+e^y\\ & \therefore \frac{dy}{dx}(x{e}^y-1)=-e^y\Rightarrow \frac{dy}{dx}=\frac{-e^y}{x{e}^y-1}\end{array}$

From (1) and (2), we write

$\frac{dy}{dx}=-\left(\frac{y}{x}\right)\times \frac{1}{y-1}=\frac{-y}{x(y-1)}=\frac{y}{x(1-y)}$