# Derivative of x^sin(x)

noelo2014
This is not homework it's just an equation I pulled out of the air and have been trying to solve

Derivative of x^sin(x)

I know it has something to do with the "general power rule" but I cannot figure it out, I'd really love someone to explain it to me conceptually instead of just showing me some trick or short cut to solving it. Thanks ever so much!

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2022 Award
Hint: Write (for $x>0$)
$$x^{\sin x}=\exp(\ln x \cdot \sin x).$$

Tanya Sharma
Hi noelo2014...

The general approach to find derivative of functions of the form y=g(x)h(x) is to first take log on both the sides and then differentiate.

y=g(x)h(x)
logy = log{g(x)h(x)}
logy = h(x)log{g(x)}
(1/y)y' = h'(x)log{g(x)} + {h(x)/g(x)}g'(x)
y' = g(x)h(x)[h'(x)log{g(x)} + {h(x)/g(x)}g'(x)]

Hope this helps

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Dearly Missed
I'd rather say that is a neat trick, Tanya Sharma, rather than a "general rule". The general rule is application of the chain rule in some manner.
Here's a way to make use of the multi-variable chain rule, for a change.
Consider the function $g(x,y)=x^{y}, Y(x)=\sin(x),\to{f}(x)=g(x,Y(x))$
Thus, we easily have:
$$\frac{df}{dx}=\frac{\partial{g}}{\partial{x}}+ \frac{\partial{g}}{\partial{x}}\frac{dY}{dx}$$

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noelo2014
I'd rather say that is a neat trick, Tanya Sharma, rather than a "general rule". The general rule is application of the chain rule in some manner.
Here's a way to make use of the multi-variable chain rule, for a change.
Consider the function $g(x,y)=x^{y}, Y(x)=\sin(x),\to{f}(x)=g(x,Y(x))$
Thus, we easily have:

I think there should be a ∂y/∂x there, I think it should be

$$\frac{df}{dx}=\frac{\partial{g}}{\partial{x}}+ \frac{\partial{g}}{\partial{y}}\frac{dY}{dx}$$

And thanks...

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Aargh, yes. stevmg
So, what's the answer for the first derivative of x^sin x ?

[(1/x)sin x + (ln x)cos x](x^sin x)?

Last edited:
Gold Member
2022 Award
Just use
$$f(x)=x^{\sin x}=\exp(\sin x \ln x)$$
which holds for $x>0$ (as a real function). Then you simply take the derivative, using the chain and product rule:
$$f'(x)=\exp(\sin x \ln x) (\sin x \ln x)'=\exp(\sin x \ln x) \left (\cos x \ln x+\frac{\sin x}{x} \right )=x^{\sin x} \left (\cos x \ln x+\frac{\sin x}{x} \right ).$$

stevmg
vanhees71 -

Guess what? That's what I got (look at my post just above yours)

I guess I am not as senile as I thought.

Thanks for your input and help.