http://i48.tinypic.com/2s1165i.jpg

so the answer is s/b (should be) on lower right, no idea how they got there, anyone know where I went wrong, thanks?

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- Feb 5th 2013, 01:51 PMtogotrying to get right answer on derivative
http://i48.tinypic.com/2s1165i.jpg

so the answer is s/b (should be) on lower right, no idea how they got there, anyone know where I went wrong, thanks? - Feb 5th 2013, 02:57 PMPlatoRe: trying to get right answer on derivative
- Feb 5th 2013, 04:00 PMProve ItRe: trying to get right answer on derivative
- Feb 5th 2013, 08:56 PMhollywoodRe: trying to get right answer on derivative
The problem seems to be to take the derivative of $\displaystyle \ln(e\sin{x^2})$, for which the correct solution is $\displaystyle \frac{2x}{\tan{x^2}}$. That's e times $\displaystyle \sin{x^2}$ in parentheses.

I don't follow your solution starting at the second line - it looks like it should be $\displaystyle v=e\sin{x^2}$ (not $\displaystyle v'$), and I don't get the stuff in brackets at all.

The solution to this derivative is pretty straightforward - just use the chain rule and work from the outside in:

$\displaystyle \frac{d}{dx} \ln(e\sin{x^2})$: the derivative of $\displaystyle \ln{u}$ is $\displaystyle \frac{1}{u}$, so use the chain rule

$\displaystyle \frac{1}{e\sin{x^2}} \frac{d}{dx} e\sin{x^2}$: bring out the constant e and cancel

$\displaystyle \frac{1}{\sin{x^2}} \frac{d}{dx} \sin{x^2}$: the derivative of $\displaystyle \sin{u}$ is $\displaystyle \cos{u}$, so use the chain rule

$\displaystyle \frac{1}{\sin{x^2}} \cos{x^2} \frac{d}{dx} x^2$: the derivative of $\displaystyle x^2$ is $\displaystyle 2x$

$\displaystyle \frac{1}{\sin{x^2}} (\cos{x^2})( 2x)$

$\displaystyle \frac{2x}{\tan{x^2}}$

- Hollywood