Question

Suppose, $A=\frac{dy}{dx}$ of $x^2+y^2=4$ at $(\sqrt{2},\sqrt{2}),B=\frac{dy}{dx}$ of $siny+sinx=sinx\cdot siny$ at $(\pi ,\pi )$ and $C=\frac{dy}{dx}$ of $2e^{xy}+e^x{e}^y-e^x=e^{xy+1}$ at $(1,1)$, then $(A-B-C)$ has the value equal to .....

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

$\frac{1}{2}$

$A:\frac{d}{dx}(x^2+y^2=4)$ at $(\sqrt{2},\sqrt{2}))$
$2x+2y\frac{dy}{dx}=0$
$\frac{dy}{dx}=-\frac{x}{y}=-\frac{\sqrt{2}}{\sqrt{2}}=-1$
$B:\frac{d}{dx}(\sin y+\sin x=\sin x\cdot \sin y)$ at $(\pi ,\pi )$
$\cos y\frac{dy}{dx}+\cos x=\sin x\cos y\frac{dy}{dx}+\sin y\cos x$
$\frac{dy}{dx}(\cos y-\sin x\cos y)=\sin y\cos x-\cos x$
$\frac{dy}{dx}=\frac{\cos x\left(\sin y-1\right)}{\cos y(1-\sin x)}$
$B=\frac{-1(0-1)}{-1(1-0)}=-1$
$C:\frac{d}{dx}(2e^{xy}+e^x{e}^y-e^x=e^{xy+1})$ at $(1,1)$
$\left[2e^{xy}\left(x\frac{dy}{dx}+y\right)\right]+e^x{e}^y\frac{dy}{dx}+e^y{e}^x-e^x=e^{xy+1}\left(x\frac{dy}{dx}+y\right)$
$\frac{dy}{dx}=\frac{y{e}^{xy+1}-2y{e}^{xy}-e^{x+y}+e^x}{2x{e}^{xy}+e^{x+y}-x{e}^{xy+1}}$.......... since $e^x{e}^y=e^{x+y}$
$C=\frac{e^2-2e^1-e^2+e^1}{2e^1+e^2-e^2}$
$=\frac{-e^1}{2e^1}=-\frac{1}{2}$
Therefore $A-B-C=\frac{1}{2}$