Identify the components for the product rule.

Question

ScanSolve AI Tutor · Accepted Answer

Here's the solution to the differentiation problem:

We are given the function $f(x) = x^2\sqrt{1-x^2}$ and need to show that $f'(x) = \frac{x(2-3x^2)}{\sqrt{1-x^2}}$ .

Identify the components for the product rule.
Let $u = x^2$ and $v = \sqrt{1-x^2} = (1-x^2)^{1/2}$ .
Differentiate $u$ and $v$ .
Differentiating $u$ :
$\frac{du}{dx} = 2x$
Differentiating $v$ using the chain rule:
$\frac{dv}{dx} = \frac{1}{2}(1-x^2)^{-1/2} \cdot (-2x)$ $\frac{dv}{dx} = -x(1-x^2)^{-1/2}$ $\frac{dv}{dx} = \frac{-x}{\sqrt{1-x^2}}$
Apply the product rule, $f'(x) = u\frac{dv}{dx} + v\frac{du}{dx}$ .
$f'(x) = (x^2)\left(\frac{-x}{\sqrt{1-x^2}}\right) + (\sqrt{1-x^2})(2x)$ $f'(x) = \frac{-x^3}{\sqrt{1-x^2}} + 2x\sqrt{1-x^2}$
Combine the terms by finding a common denominator.
The common denominator is $\sqrt{1-x^2}$ .
$f'(x) = \frac{-x^3}{\sqrt{1-x^2}} + \frac{2x\sqrt{1-x^2} \cdot \sqrt{1-x^2}}{\sqrt{1-x^2}}$ $f'(x) = \frac{-x^3 + 2x(1-x^2)}{\sqrt{1-x^2}}$
Simplify the numerator.
$f'(x) = \frac{-x^3 + 2x - 2x^3}{\sqrt{1-x^2}}$ $f'(x) = \frac{2x - 3x^3}{\sqrt{1-x^2}}$
Factor out $x$ from the numerator.
$f'(x) = \frac{x(2 - 3x^2)}{\sqrt{1-x^2}}$
This matches the expression we needed to show.

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