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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.

Michael Hardy
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user3658777
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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.

k170
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$$y = e^{\ln (\ln(x))^{\cos x}} = e^{\cos x \ln (\ln(x))}$$

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

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

math student
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    $$e^{\cos\left(x\right)\log\left|\log\left|x\right|\right|}\implies \log\left|x\right|^{\cos\left(x\right)} \left(-\sin\left(x\right)\log\left| \log\left|x\right|\right|+\cos\left(x\right)\frac{1}{x\log\left|x\right|}\right)$$ – bjd2385 Oct 18 '14 at 02:53
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    you re right . thanks! i will fix the error – math student Oct 18 '14 at 04:14
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I will assume you mean $(\ln x)^{\cos x}$ and not $\ln(x^{\cos x})$. I wish people would be explicit about that point.

If $g(x)$ is constant then $\dfrac d {dx} (f(x))^{g(x)} = g(x)f(x)^{g(x)-1}$.

If $f(x)$ is constant then $\dfrac d {dx} (g(x))^{g(x)} = f(g)^{g(x)} (\ln f(x)) f'(x)$.

If neither is constant then $\dfrac d {dx} (g(x))^{g(x)} = \text{the sum of those two things}$.

That can be proved by logarithmic differentiation.

Michael Hardy
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