Given {eq}f(x) = (x^2 - 2x - 1)^{2/3} {/eq}, find {eq}f'(0) {/eq}


Given {eq}f(x) = (x^2 - 2x - 1)^{2/3} {/eq}, find {eq}f'(0) {/eq}

The chain rule:

If the function {eq}g {/eq} is differentiable in {eq}x {/eq} and the function {eq}f {/eq} in {eq}g(x) {/eq}, then the composite function {eq}f \circ g {/eq} is differentiable in {eq}x {/eq} and {eq}(f \circ g)^\prime(x)=f^\prime(g(x))g^\prime(x) {/eq}.

Answer and Explanation: 1

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Function: {eq}f(x) = (x^2 - 2x - 1)^{2/3} {/eq}

Using the chain rule:

{eq}f^\prime(x) =\frac{2}{3} (x^2 - 2x -...

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Learn more about this topic:

The Chain Rule for Partial Derivatives


Chapter 14 / Lesson 4

This lesson defines the chain rule. It goes on to explore the chain rule with partial derivatives and integrals of partial derivatives.

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