# 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.

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.