# Derivative of a Function at a point

• Nov 8th 2009, 12:41 PM
StarlitxSunshine
Derivative of a Function at a point
If $xsiny +ysinx = 0$, then, $\frac{dy}{dx}$ at $\frac{\pi}{4},\frac{\pi}{4}$ = ?

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

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

Using implicit differentiation:

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

$
xcosy + siny +ycosx + sinyy' 0
$

$
sinyy' = -ycosx-siny-xcosy
$

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

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

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

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

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

$
y'= -91$
• Nov 8th 2009, 12:48 PM
I think you did the implicit differentiation wrong. Here's what I obtained:

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

Substitute these values in the expression above:

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

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

So now we have:

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

Try solving for $\frac{dy}{dx}$ and see if you can get an answer.
• Nov 8th 2009, 12:59 PM
VonNemo19
Quote:

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

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

Using implicit differentiation:

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

$
xcosy + siny +ycosx + sinyy' 0
$

$
sinyy' = -ycosx-siny-xcosy
$

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

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

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

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

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

$
y'= -91$

You must understand that y is a function of x. Therefore whenevever you take the derivative of y, treat is as such...

$(xcosy\frac{dy}{dx}+siny)+(ycosx+\frac{dy}{dx}siny )=0$
• Nov 8th 2009, 01:08 PM
StarlitxSunshine
Yup Yup :3