Why isn't $\frac{d}{dx}(x^{sinx})$ simply $\frac{d}{du}(x^{u})*\frac{d}{dx}(sinx)$?

Question

I know that this solution is wrong. The correct one is $\frac{d}{dx}(x^{\sin x}) = x^{\sin x - 1} (\sin x + x \log x \cos x)$, however I can't catch why there is a sum as result of the chain rule. I've seen this derivation with a log and an exponential trick but I'm not convinced yet. I didn't find any direct derivation without these two tricks. It would be of great help if someone did that here.

Answer 1

When you are applying your chain rule, you are forgetting to differentiate the $x$ term in $x^{(\cdots)}$.

Your derivative is easier to calculate if you set $x = e^{\ln(x)}$. In this case, you have: $$x^{\sin(x)} = e^{\ln(x)\sin(x)}$$ and the derivative is simply: $$\frac{d}{dx}x^{\sin(x)} = e^{\ln(x)\sin(x)}\bigg{[}\frac{1}{x}\sin(x)+\ln(x)\cos(x)\bigg{]} = x^{\sin(x)-1}\bigg{[}\sin(x)+x\ln(x)\cos(x)\bigg{]}$$

Answer 2

$x$ is a function of $u$ so the first factor has to be $$\frac d{du}\left((\arcsin u)^u\right)$$

Answer 3

The trouble with differentiating things like $f(x)^{g(x)}$ is that neither of the rules for handling things like $f(x)^c$ or $c^{f(x)}$ will apply. Neither the base or the exponents are constants, this is a uniquely different situation. The 'trick' for these things are to use logarithms and implicit differentiation. What we do to handle this situation is the following: \begin{align*} y &= f(x)^{g(x)} \implies \ln (y) = f(x)\ln g(x) \implies y'/y = f'(x) \ln(g(x)) + f(x) \frac{g'(x)}{g(x)}\\ y' &= yf'(x) \ln(g(x)) + yf(x) \frac{g'(x)}{g(x)} = f(x)^{g(x)}f'(x) \ln(g(x)) + f(x)^{g(x)}f(x) \frac{g'(x)}{g(x)} \end{align*}

Replacing $f$ and $g$ with $\sin(x)$ and $x$ will give you the result you would expect.