1. ## derivative help

I'm having trouble figuring out how to find the derivative of xcosx. I know the answer is xcosx (-sinx lnx + (cosx)/x) but i have no idea how to get there.

2. ## Re: derivative help

Originally Posted by JML2618
I'm having trouble figuring out how to find the derivative of xcosx. I know the answer is xcosx (-sinx lnx + (cosx)/x) but i have no idea how to get there.
$x^{y}=e^{ln[x]y}$

see if that gets you started

3. ## Re: derivative help

Originally Posted by JML2618
I'm having trouble figuring out how to find the derivative of xcosx. I know the answer is xcosx (-sinx lnx + (cosx)/x) but i have no idea how to get there.
I also encouraged student to study the following and be able to tell me by way of logarithmic differentiation why it is true.

Suppose that each of $f~\&~g$ is a differential function, then
${D_x}\left( {g{{(x)}^{f(x)}}} \right) = \left( {g{{(x)}^{f(x)}}} \right)\left( {f'(x)\log (g(x)) + f(x)\frac{{g'(x)}}{{g(x)}}} \right)$

4. ## Re: derivative help

I would have done it as "if $y= x^{cos(x)}$ then $ln(y)= cos(x) ln(x)$ so that $\frac{1}{y}y'= -sin(x)ln(x)+ \frac{cos(x)}{x}$ and then
$y'= y\left(-sin(x)ln(x)+ \frac{cos(x)}{x}\right)= x^{cos(x)}\left(-sin(x)ln(x)+ \frac{cos(x)}{x}\right)$".

5. ## Re: derivative help

Thanks for the replies they helped and i understand it now!