I'm having trouble figuring out how to find the derivative of x^{cosx}. I know the answer is x^{cosx }(-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 $\displaystyle f~\&~g$ is a differential function, then
$\displaystyle {D_x}\left( {g{{(x)}^{f(x)}}} \right) = \left( {g{{(x)}^{f(x)}}} \right)\left( {f'(x)\log (g(x)) + f(x)\frac{{g'(x)}}{{g(x)}}} \right)$
I would have done it as "if $\displaystyle y= x^{cos(x)}$ then $\displaystyle ln(y)= cos(x) ln(x)$ so that $\displaystyle \frac{1}{y}y'= -sin(x)ln(x)+ \frac{cos(x)}{x}$ and then
$\displaystyle y'= y\left(-sin(x)ln(x)+ \frac{cos(x)}{x}\right)= x^{cos(x)}\left(-sin(x)ln(x)+ \frac{cos(x)}{x}\right)$".