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.

$$y = e^{\ln (\ln(x))^{\cos x}} = e^{\cos x \ln (\ln(x))}$$
$$y^{'} = e^{\cos x \ln (\ln(x))} \cdot (\cos x \ln (\ln(x)))^{'} = e^{\cos x \ln (\ln(x))}  \cdot \left(-\sin x \ln (\ln(x)) + \cos x \frac{1}{\ln x} \frac{1}{x}\right)$$
$$=  (\ln(x))^{\cos x}\left(-\sin x \ln (\ln(x)) + \cos x \frac{1}{\ln x} \frac{1}{x}\right)$$
