# Derivative of f(x)=(square root x^2-2x)^3-9(square rootx^2-2x)

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

mathwonk
Homework Helper
2020 Award
use the prower rule and the chain rule. they are in your book or online.

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

Just like what Mathwonk said, use the chain rule for the first one
$$(\sqrt{x^2 - 2x})^3$$

$$\sqrt{x^2 -2x}$$ is the inside function g(x) and $$x^3$$ is the outer function f(x)

The chain rule states, $$h '(x) = f '(g(x)) * g'(x)$$

For the power rule, for
$$f(x)g(x)$$, the derivative is $$f'(x)g(x) + g'(x)f(x)$$

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:
PhysicsinCalifornia said:
For the power rule, for
$$f(x)g(x)$$, the derivative is $$f'(x)g(x) + g'(x)f(x)$$
That's the product rule. Power rule:

$$\frac{d}{dx} x^n = nx^{n - 1}$$

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