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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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.Quote:
Originally Posted by togo
I doubt anyone can.
Why don't you type it out?
When you write $\displaystyle \displaystyle \begin{align*} e\,\sin{\left( x^2 \right)} \end{align*}$ do you mean $\displaystyle \displaystyle \begin{align*} e^{\sin{\left( x^2 \right)}} \end{align*}$?Quote:
Originally Posted by togo
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
