How do you find the derivative of #y=3/4(x^(2)-1)^(2/3)#?

Question

Christopher Allers · Accepted Answer

#f'(x)=x/root(3)(x^2-1)#

#f(x)=3/4(x^2-1)^(2/3)#

Derivative: #f'(x)=3/4*2/3*(x^2-1)^(-1/3)*2x#

#color(blue)(f'(x)=x/root(3)(x^2-1))#

Steps:

#1" "# #3/4# is a constant so we can bring that out.

#2" "# Using Power Rule, we can get #2/3(x^2-1)^(-1/3)#. However, since #(x^2-1)# isn't #x#, we have to use Chain Rule.

#3" "# We get the derivative of #(x^2-1)# which is #2x# and multiply it.

Sorry, I'm not very good at explaining this. Just let me know if you have questions.

Evelyn Agard · Answer

undefined To find the derivative of $ y = \frac{3}{4}(x^2 - 1)^{2/3} $, you can use the chain rule.

First, differentiate the outer function with respect to the inner function:

$ \frac{dy}{du} = \frac{2}{3}(x^2 - 1)^{-1/3} \cdot 2x $

Then, differentiate the inner function:

$ \frac{du}{dx} = 2x $

Now, apply the chain rule:

$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{2}{3}(x^2 - 1)^{-1/3} \cdot 2x \cdot 2x $

Simplify:

$ \frac{dy}{dx} = \frac{4x(x^2 - 1)^{-1/3}}{3} $