# Thread: Find The Derivative of the Function x^2√(1-x^2)

1. ## Find The Derivative of the Function x^2√(1-x^2)

f(x)=(x^2)(√(1-x^2))

f'(x)=(2x)(√(1-x^2))+(x^2)(1/2)(√(1-x^2))^-1)(-2x)

f'(x)=(2x)((1-x^2)^1/2)+(x^2)(1/2)(√(1-x^2)^-1)(-2x)

f'(x)=√(4x^2-4x^4)+(1/2)(-2x^3)(1/(√(1-x^2))

umm i'm lost, i probably didn't even get the above right (sorry the format sucks pretty bad)

never mind i found a picture

2. ## Re: Find The Derivative of the Function x^2√(1-x^2)

$\displaystyle x^2\sqrt{1-x^2}$

I'd use the product rule.

Let $\displaystyle u=x^2$

then $\displaystyle \frac{du}{dx}=?$

Let $\displaystyle v=(1-x^2)^{\frac{1}{2}}$

then $\displaystyle \frac{dv}{dx}=?$

You should be able to proceed from here.

3. ## Re: Find The Derivative of the Function x^2√(1-x^2)

$\displaystyle x^2\sqrt{1-x^2}=\sqrt{x^4-x^6}$

and

$\displaystyle \frac{d}{dx}(\sqrt{f(x)})=\frac{f'(x)}{2\sqrt{f(x) }}$

4. ## Re: Find The Derivative of the Function x^2√(1-x^2)

Originally Posted by Deo3560
f(x)=(x^2)(√(1-x^2))

f'(x)=(2x)(√(1-x^2))+(x^2)(1/2)(√(1-x^2))^-1)(-2x)

f'(x)=(2x)((1-x^2)^1/2)+(x^2)(1/2)(√(1-x^2)^-1)(-2x)
$\displaystyle f'(x)=(2x)(1-x^2)^{1/2}-(x^2)(1/2)(1-x^2)^{-1/2}(-2x)$

It is right up to here, and may be simplified:

$\displaystyle f'(x)=(2x)(1-x^2)^{1/2}+\frac{x^3}{(1-x^2)^{1/2}}$

It is easier to proceed by taking out a factor of $\displaystyle x \sqrt{1-x^2)$

$\displaystyle f'(x) =x(1-x^2)^{1/2}\left[2 -\frac{x^2}{1-x^2} \right]$

That square bracket may be simplified further if you wish.

CB