Differentiation involving Sin(x) as a power

Homework Statement

Find the derivative of y=(x^2)^sinx; using the chain rule.

Homework Equations

No other relevant equations.

The Attempt at a Solution

I attempted to apply the Chain rule: dy/dx = dy/du X du/dx
Subbing u for x^2, which made y = u^sinx

I ended up with du/dx = 2x;

and dy/du = ln(u)*cos(x)*(U)^sinx

I know this isn't right, and all i've spent the past hour doing is confusing myself. I've searched my text book, and online, and can't find anyting (that I understand) to help me.
To be honest, i'm not even sure how i'm supposed to apply the chain rule

Any help would be appreciated, thank you.

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Mute
Homework Helper
The problem with the way you have attempted it is that when you computed dy/du you did not write y entirely in terms of u, you left it in terms of u and x, but u and x are not independent. This is why your calculation isn't working you. You should replace x with ##\sqrt{u}## to do it properly.

There is a slightly easier way to do this, as well. Note that you can write

##x^2 = e^{\ln (x^2)} = e^{2\ln |x|}.##

Then, recall that ##(a^b)^c = a^{bc}## for real numbers a, b and c. This will give you a form that is easier to differentiate. (You'll want to use a different chain rule substitution, though).

SammyS
Staff Emeritus
Homework Helper
Gold Member

Homework Statement

Find the derivative of y=(x^2)^sinx; using the chain rule.

Homework Equations

No other relevant equations.

The Attempt at a Solution

I attempted to apply the Chain rule: dy/dx = dy/du X du/dx
Subbing u for x^2, which made y = u^sinx

I ended up with du/dx = 2x;

and dy/du = ln(u)*cos(x)*(U)^sinx

I know this isn't right, and all i've spent the past hour doing is confusing myself. I've searched my text book, and online, and can't find anyting (that I understand) to help me.
To be honest, I'm not even sure how I'm supposed to apply the chain rule

Any help would be appreciated, thank you.
I can see a couple of ways to do this with the chain rule. One uses the multi-variable chain rule. The other require rewrite the function.

Multi-variable chain rule:
Let u(x)= x2 and v(x) = sin(x)

Then $y=f(x,\,y)=u^v\,.$

Then use $\displaystyle \frac{dy}{dx}=\frac{\partial y}{\partial u}\frac{du}{dx}+\frac{\partial y}{\partial v}\frac{dv}{dx}\ .$

The single variable method:
$\displaystyle y=(x^2)^{\sin(x)}=e^{2\ln(x)\sin(x)}\ .$

Now apply the chain rule. You will also need the product rule.​

HallsofIvy
Simpler, I think, than writing $f(x)^{g(x)}$ as $e^{g(x)ln(f(x))}$ is to start with $y(x)= f(x)^{g(x)}$ and take the logarithm of both sides: ln(y)= g(x)ln(f(x)).
Here, the function is $y(x)= (x^2)^{sin(x)}= x^{2sin(x)}$. Then $ln(y)= 2sin(x) ln(x)$. Then you can use the product rule to differentiate the right side using the product rule. Of course, the derivative of ln(y), on the left, is (1/y)dy/dx. After taking the derivative of both sides multiply through by $y= (x^2)^{sin(x)}$ to find dy/dx.