if y=x^cosx, find dy/dx.
help please...Im not sure if I did it right.
Do you HAVE to use logarithmic differentiation? You can just use chain rule...
$\displaystyle y = x^{\cos{(x)}}$.
Let $\displaystyle u = \cos{(x)}$ so that $\displaystyle y = x^u$.
$\displaystyle \frac{du}{dx} = -\sin{(x)}$
$\displaystyle \frac{dy}{du} = ux^{u - 1} = \cos{(x)}\,x^{\cos{(x)}-1}$
So $\displaystyle \frac{dy}{dx} = -\sin{(x)}\cos{(x)}\,x^{\cos{(x)}-1}$.