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How do I find the derivative of f(x)=(square root x^2-2x)^3-9(square rootx^2-2x)

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How do I find the derivative of f(x)=(square root x^2-2x)^3-9(square rootx^2-2x)

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mathwonk

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use the prower rule and the chain rule. they are in your book or online.

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[tex]f(x) = (\sqrt{x^2 - 2x})^3 - 9\sqrt{x^2 - 2x}[/tex]candynrg said:How do I find the derivative of f(x)=(square root x^2-2x)^3-9(square rootx^2-2x)

Just like what Mathwonk said, use the chain rule for the first one

[tex] (\sqrt{x^2 - 2x})^3[/tex]

[tex] \sqrt{x^2 -2x} [/tex] is the inside function g(x) and [tex] x^3 [/tex] is the outer function f(x)

The chain rule states, [tex] h '(x) = f '(g(x)) * g'(x) [/tex]

For the power rule, for

[tex]f(x)g(x) [/tex], the derivative is [tex]f'(x)g(x) + g'(x)f(x)[/tex]

I'm giving you this information because you can find it in your book. Its up to you to find those inside and outer functions and differentiating them

Last edited:

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That's the product rule. Power rule:PhysicsinCalifornia said:For the power rule, for

[tex]f(x)g(x) [/tex], the derivative is [tex]f'(x)g(x) + g'(x)f(x)[/tex]

[tex]\frac{d}{dx} x^n = nx^{n - 1}[/tex]

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HallsofIvy

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Of course, to use the power rule you will need to know that [tex]\sqrt{x}= x^{\frac{1}{2}}[/tex].

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