# trying to get right answer on derivative

• February 5th 2013, 01:51 PM
togo
trying 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?
• February 5th 2013, 02:57 PM
Plato
Re: trying to get right answer on derivative
Quote:

Originally Posted by togo
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?

I cannot read that image at all.
I doubt anyone can.
Why don't you type it out?
• February 5th 2013, 04:00 PM
Prove It
Re: trying to get right answer on derivative
Quote:

Originally Posted by togo
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?

When you write \displaystyle \begin{align*} e\,\sin{\left( x^2 \right)} \end{align*} do you mean \displaystyle \begin{align*} e^{\sin{\left( x^2 \right)}} \end{align*}?
• February 5th 2013, 08:56 PM
hollywood
Re: trying to get right answer on derivative
The problem seems to be to take the derivative of $\ln(e\sin{x^2})$, for which the correct solution is $\frac{2x}{\tan{x^2}}$. That's e times $\sin{x^2}$ in parentheses.

I don't follow your solution starting at the second line - it looks like it should be $v=e\sin{x^2}$ (not $v'$), 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:

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

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

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

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

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

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

- Hollywood