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

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Answer

x(2−3x2)1−x2\frac{x(2 - 3x^2)}{\sqrt{1 - x^2}}

Graph of f(x) = x²√(1 − x²)

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Key points of f(x) = x²√(1 − x²)
Pointxy
x-intercept−10
Maximum−0.820.38
Minimum00
Maximum0.820.38
x-intercept10

Step-by-step solution

6 steps

To find the derivative of f(x)=x21−x2f(x) = x^2 \sqrt{1 - x^2}, we will use the product rule and the chain rule.

  1. Identify the parts for the product rule.

    Let u(x)=x2u(x) = x^2 and v(x)=1−x2=(1−x2)1/2v(x) = \sqrt{1 - x^2} = (1 - x^2)^{1/2}.
    The product rule states that if f(x)=u(x)v(x)f(x) = u(x)v(x), then f′(x)=u′(x)v(x)+u(x)v′(x)f'(x) = u'(x)v(x) + u(x)v'(x).

  2. Find the derivative of u(x)u(x).

    u′(x)=ddx(x2)=2xu'(x) = \frac{d}{dx}(x^2) = 2x

  3. Find the derivative of v(x)v(x) using the chain rule.

    v′(x)=ddx((1−x2)1/2)v'(x) = \frac{d}{dx}((1 - x^2)^{1/2})

    Apply the power rule and chain rule:

    v′(x)=12(1−x2)−1/2⋅ddx(1−x2)v'(x) = \frac{1}{2}(1 - x^2)^{-1/2} \cdot \frac{d}{dx}(1 - x^2) v′(x)=12(1−x2)−1/2⋅(−2x)v'(x) = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x) v′(x)=−x(1−x2)−1/2=−x1−x2v'(x) = -x(1 - x^2)^{-1/2} = \frac{-x}{\sqrt{1 - x^2}}

  4. Apply the product rule.

    f′(x)=u′(x)v(x)+u(x)v′(x)f'(x) = u'(x)v(x) + u(x)v'(x) f′(x)=(2x)(1−x2)+(x2)(−x1−x2)f'(x) = (2x)(\sqrt{1 - x^2}) + (x^2)\left(\frac{-x}{\sqrt{1 - x^2}}\right) f′(x)=2x1−x2−x31−x2f'(x) = 2x\sqrt{1 - x^2} - \frac{x^3}{\sqrt{1 - x^2}}

  5. Combine the terms by finding a common denominator.

    The common denominator is 1−x2\sqrt{1 - x^2}.

    f′(x)=2x1−x2⋅1−x21−x2−x31−x2f'(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)=2x(1−x2)−x31−x2f'(x) = \frac{2x(1 - x^2) - x^3}{\sqrt{1 - x^2}} f′(x)=2x−2x3−x31−x2f'(x) = \frac{2x - 2x^3 - x^3}{\sqrt{1 - x^2}} f′(x)=2x−3x31−x2f'(x) = \frac{2x - 3x^3}{\sqrt{1 - x^2}}

  6. Factor the numerator.

    f′(x)=x(2−3x2)1−x2f'(x) = \frac{x(2 - 3x^2)}{\sqrt{1 - x^2}}

    The final answer is x(2−3x2)1−x2\frac{x(2 - 3x^2)}{\sqrt{1 - x^2}}.

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