# Why can't I differentiate $x^{\sin x}$ using the power rule?

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 $$\cos x$$.

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 $$(\sin x\ln x)'\ne \frac{\sin x}x$$.

• Does that proof hold for $k \lt 0$? – user150203 Jan 6 '19 at 11:44
• 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 $x \gt 0$ ? – user150203 Jan 6 '19 at 11:48
• No, I appreciate that - but just for clarity - this method as presented only holds for $x \gt 0$? – user150203 Jan 6 '19 at 11:50
• 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 $x \gt 0$? – user150203 Jan 6 '19 at 11:53
• @DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P – TheSimpliFire Jan 6 '19 at 11:54

$$(x^k)'=kx^{k-1},$$ 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?

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

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 $$\sin x$$.

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