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I'm trying to differentiate $ x^{\sin x} $, with respect to $ x $ and $x > 0 $. My textbook initiates with $ y = x^{\sin x} $, takes logarithms on both sides and arrives at the answer $$ x^{\sin x - 1}.\sin x + x^{\sin x}.\cos x \ \log x $$

Why can't I use the power rule to proceed like this: $ y' = (\sin x) \ x^{\sin x-1} \cos x $.

Here, I've first differentiated $ x $ with respect to $ \sin x $ and then I've differentiated $ \sin x $ with respect to $ x $ to get $ \cos x $.

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The power rule states that the derivative of a function of the form $x^k$ is $kx^{k-1}$ where $k$ is a real number. In your case, $\sin x$ is variable, and can takes values from $-1$ to $+1$, so is not constant.


To prove the power rule, note that $y=x^k\implies \ln y=k\ln x\implies \frac{y'}y=\frac kx\cdot$ via the chain rule. Therefore, $y'=x^k\frac kx=kx^{k-1}$. When $K=\sin x$, we need to do some more to manipulate the RHS, since $(\sin x\ln x)'\ne \frac{\sin x}x$.

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  • $\begingroup$ Does that proof hold for $k \lt 0$? $\endgroup$ – user150203 Jan 6 '19 at 11:44
  • $\begingroup$ Sorry, my apologies, I meant to type what if $x^k \lt 0$ - can you still take the logarithm? If not, does this then only hold for $x \gt 0$ ? $\endgroup$ – user150203 Jan 6 '19 at 11:48
  • $\begingroup$ No, I appreciate that - but just for clarity - this method as presented only holds for $x \gt 0$? $\endgroup$ – user150203 Jan 6 '19 at 11:50
  • $\begingroup$ I agree, I'm not disputing that. Nor any I challenging that the derivative as you find holds for $x \lt 0$. I'm just curious, in what you have presented in the solution - does it hold for all $x$? or is the reasoning (again as presented) only applicable for $x \gt 0$? $\endgroup$ – user150203 Jan 6 '19 at 11:53
  • $\begingroup$ @DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P $\endgroup$ – TheSimpliFire Jan 6 '19 at 11:54
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$(x^k)'=kx^{k-1},$ where $k$ is constant.

In our case $\sin{x}\neq constant$ and we can not use this rule.

By the way, $$\left(x^{\sin{x}}\right)'=\left(e^{\sin{x}\ln{x}}\right)'=e^{\sin{x}\ln{x}}(\sin{x}\ln{x})'=...$$ Can you end it now?

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Because the power rule requires that the exponent be a constant. $\sin x$ is most definitely not a constant.

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You are trying to use the rule $$\frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$$

But this rule doesn't apply here, because $x^{\sin x}$ is not just a function of $\sin x$, it is a function of $x$ and $\sin x$.

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  • $\begingroup$ Sometimes we can write $x=\arcsin\sin{x}$ and we'll obtain the function of $\sin{x}.$ :) $\endgroup$ – Michael Rozenberg Dec 25 '18 at 11:49

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