f(x)=x2 * sqrt(1-x2) find f'(x)

To find the derivative of $f(x) = x^2 \sqrt{1 - x^2}$, we will use the product rule and the chain rule.

Step 1: Identify the parts for the product rule. Let $u(x) = x^2$ and $v(x) = \sqrt{1 - x^2} = (1 - x^2)^{1/2}$. The product rule states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

Step 2: Find the derivative of $u(x)$. $u'(x) = \frac{d}{dx}(x^2) = 2x$

Step 3: Find the derivative of $v(x)$ using the chain rule. $v'(x) = \frac{d}{dx}((1 - x^2)^{1/2})$ Apply the power rule and chain rule: $v'(x) = \frac{1}{2}(1 - x^2)^{-1/2} \cdot \frac{d}{dx}(1 - x^2)$ $v'(x) = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x)$ $v'(x) = -x(1 - x^2)^{-1/2} = \frac{-x}{\sqrt{1 - x^2}}$

Step 4: Apply the product rule. $f'(x) = u'(x)v(x) + u(x)v'(x)$ $f'(x) = (2x)(\sqrt{1 - x^2}) + (x^2)\left(\frac{-x}{\sqrt{1 - x^2}}\right)$ $f'(x) = 2x\sqrt{1 - x^2} - \frac{x^3}{\sqrt{1 - x^2}}$

Step 5: Combine the terms by finding a common denominator. The common denominator is $\sqrt{1 - x^2}$. $f'(x) = \frac{2x\sqrt{1 - x^2} \cdot \sqrt{1 - x^2}}{\sqrt{1 - x^2}} - \frac{x^3}{\sqrt{1 - x^2}}$ $f'(x) = \frac{2x(1 - x^2) - x^3}{\sqrt{1 - x^2}}$ $f'(x) = \frac{2x - 2x^3 - x^3}{\sqrt{1 - x^2}}$ $f'(x) = \frac{2x - 3x^3}{\sqrt{1 - x^2}}$

Step 6: Factor the numerator. $f'(x) = \frac{x(2 - 3x^2)}{\sqrt{1 - x^2}}$

The final answer is $\boxed{\frac{x(2 - 3x^2)}{\sqrt{1 - x^2}}}$.

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