# Thread: Derivative of a Function at a point

1. ## Derivative of a Function at a point

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

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

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

Using implicit differentiation:

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

$\displaystyle xcosy + siny +ycosx + sinyy' 0$

$\displaystyle sinyy' = -ycosx-siny-xcosy$

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

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

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

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

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

$\displaystyle y'= -91$

2. I think you did the implicit differentiation wrong. Here's what I obtained:

$\displaystyle \frac{d}{dx}[xsin(y)+ysin(x)]=\frac{dx}{dx}sin(y)+x\frac{d}{dx}sin(y)+\frac{dy} {dx}sin(x)+y\frac{d}{dx}sin(x)$

Substitute these values in the expression above:

$\displaystyle \frac{d}{dx}sin(x)=cos(x)$

$\displaystyle \frac{d}{dx}sin(y)=\frac{dy}{dx}cos(y)$

So now we have:

$\displaystyle sin(y)+x\frac{dy}{dx}cos(y)+\frac{dy}{dx}sin(x)+yc os(x)=0$

Try solving for $\displaystyle \frac{dy}{dx}$ and see if you can get an answer.

3. Originally Posted by StarlitxSunshine
If $\displaystyle xsiny +ysinx = 0$, then, $\displaystyle \frac{dy}{dx}$ at $\displaystyle \frac{\pi}{4},\frac{\pi}{4}$ = ?

1. (A) -1
2. (B) 1
3. (C) 0
4. (D) $\displaystyle \frac{\pi}{4}$
5. (E)$\displaystyle \frac{-\pi}{4}$
I think I made a mistake somewhere in my working. Here's what I did:

Using implicit differentiation:

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

$\displaystyle xcosy + siny +ycosx + sinyy' 0$

$\displaystyle sinyy' = -ycosx-siny-xcosy$

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

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

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

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

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

$\displaystyle y'= -91$
You must understand that y is a function of x. Therefore whenevever you take the derivative of y, treat is as such...

$\displaystyle (xcosy\frac{dy}{dx}+siny)+(ycosx+\frac{dy}{dx}siny )=0$

4. Yup Yup :3