Find the Derivative of f(x)= 9/4 (x^2-1)2/3

[Hckyplayer8]

It's a bit regressive but after going a couple weeks without having to use the chain rule, I've gotten rusty.

The chain rule is f'(g(x)) times g'(x).

So first the outside function

[(9/4)(2/3)] (x²-1)^{2/3 - 1} = 3/2(x²-1)^-1/3

The inside function is (x² - 1) which has the derivative of 2x.

So now I do what?

 

[pka]

It's a bit regressive but after going a couple weeks without having to use the chain rule, I've gotten rusty.
The chain rule is f'(g(x)) times g'(x).
So first the outside function
[(9/4)(2/3)] (x²-1)^{2/3 - 1} = 3/2(x²-1)^-1/3
The inside function is (x² - 1) which has the derivative of 2x.

The derivative of \(\displaystyle f\circ h\circ g(x)=f'(h\circ g(x))h'(g(x))g'(x)\)

 

[Hckyplayer8]

So instead it is (9/4)(2/3u^1/3) (2x) which works out to be 3x/(x²-1)^1/3 ?

 

[Steven G]

It's a bit regressive but after going a couple weeks without having to use the chain rule, I've gotten rusty.

The chain rule is f'(g(x)) times g'(x).

So first the outside function

[(9/4)(2/3)] (x²-1)^{2/3 - 1} = 3/2(x²-1)^-1/3

The inside function is (x² - 1) which has the derivative of 2x.

So now I do what?

I take it that you are serious. You should do exactly what you said needs to be done. You said f'(g(x)) times g'(x). You found f'(g(x)) and g'(x) so multiply them.

 

[Hckyplayer8]

I take it that you are serious. You should do exactly what you said needs to be done. You said f'(g(x)) times g'(x). You found f'(g(x)) and g'(x) so multiply them.

Yes. It's the f' of g(x) that always screws me up when it comes to chain rule problems.

f'(x)...got it
g'(x)...got it

but f'(g(x))...I need some work and I was just checking to see if I have everything setup correctly.

 

[HallsofIvy]

One way to handle that is to replaced g(x) by a single letter. Here, you are taking \(\displaystyle g(x)= x^2+ 1\) and \(\displaystyle f(g(x))= \frac{9}{4}(x^2- 1)^{2/3}\) so, writing \(\displaystyle y= g(x)= x^2- 1\), we have \(\displaystyle f(y)= \frac{9}{4}y^{2/3}\) \(\displaystyle \frac{df(y)}{dy}= \frac{9}{4}\frac{2}{3}y^{2/3-1}= \frac{3}{2}y^{-1/3}\). Of course \(\displaystyle \frac{dy}{dx}= 2x\). Then, by the chain rule, \(\displaystyle \frac{df}{dx}= \frac{df}{dy}\frac{dy}{dx}= \)\(\displaystyle \frac{3}{2}y^{-1/3}\left(2x\right)= 3xy^{-1/3}\). Replacing y by \(\displaystyle x^2- 1\), \(\displaystyle \frac{df}{dx}= 3x(x^2-1)^{-1/3}\)

 

[Hckyplayer8]

Thank you for the explanation!