derivative problem. is it same? First derivative of $y=\ln(x)^{\cos x}$ is $-\sin x\ln x+\frac{\cos x}{x}$ or another answer? My friend gets another answer, but it's true? thanks.

Assuming that $\ln (x)^{\cos x}=\ln\left(x^{\cos x}\right)$, we have
$$ \frac{d}{dx}\left[\ln x^{\cos x}\right]=\frac{d}{dx}\left[(\cos x)\ln x\right]=(\cos x)\frac{d}{dx}\left[\ln x\right]+(\ln x)\frac{d}{dx}\left[\cos x\right] $$
$$ =(\cos x)\frac{1}{x}+(\ln x)(-\sin x)=\frac{\cos x}{x}-(\sin x)\ln x $$
If you meant to raise the function $\ln x$ by $\cos x$, then its best to write, $(\ln x)^{\cos x}$ as the parenthesis make it clear.  
