If $\displaystyle xsiny +ysinx = 0$, then, $\displaystyle \frac{dy}{dx}$ at $\displaystyle \frac{\pi}{4},\frac{\pi}{4}$ = ?

- (A) -1
- (B) 1
- (C) 0
- (D) $\displaystyle \frac{\pi}{4}$
- (E)$\displaystyle \frac{-\pi}{4}$

I think I made a mistake somewhere in my working. Here's what I did:

Using implicit differentiation:

$\displaystyle

\frac{dy}{dx}[xsiny +ysinx] = \frac{dy}{dx} [0]

$

$\displaystyle

xcosy + siny +ycosx + sinyy' 0

$

$\displaystyle

sinyy' = -ycosx-siny-xcosy

$

$\displaystyle

y' = \frac{-ycosx-siny-xcosy}{siny}

$

$\displaystyle

y' = \frac{\frac{-\Pi}{4}cos\frac{-\Pi}{4} - sin\frac{-\Pi}{4} -\frac{-\pi}{4}cos\frac{-\Pi}{4}}{sin\frac{-\Pi}{4}}

$

$\displaystyle

y' = \frac{ -45cos45-sin45-45cos45}{sin45}

$

$\displaystyle

y' = \frac{-45\frac{\sqrt{2}} - \frac{\sqrt{2}}{2} - 45\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}

$

$\displaystyle

y' = -91\frac{\sqrt[2]{2}}{2}*\frac{2}{\sqrt{2}}

$

$\displaystyle

y'= -91$