Derivative of Inverse Solution for f(x) = the third root of (x-8)

[Qube]

Homework Statement

If f(x) = the third root of (x-8), find the derivative of its inverse.

Homework Equations

The derivative of its inverse = 1/f'(f^-1(x)) or 1 over its derivative at its inverse.

The Attempt at a Solution

I followed both the formula to verify my solution and also did some manual work in lieu of the formula. Basically, if g is the inverse of f, then f(g(x)) = x, and so the derivative of both sides is f'(g(x)g'(x) = 1, and since g'(x) is the derivative of the inverse of x, g'(x) = 1/f'(g(x)) and I get 3x^2. I have no idea if that's correct however. Note that g(x) = f^-1'(x); I'm typing g(x) since the latter notation gets messy really quickly when you don't bother to use LATEX. :p.

https://scontent-b-mia.xx.fbcdn.net/hphotos-prn2/v/1457675_10201005261715465_14857176_n.jpg?oh=9da6ee66e6962843dfa1fcab48d3c201&oe=527B6402

 

[mfb]

There is an easy way to check your work: calculate g(x), calculate the derivative, compare it with your result.

 

[Saitama]

Your final answer is correct but you actually don't need the inverse to find its derivative. There's a simpler way.

We have
$$g(f(x))=x$$
where g(x) is the inverse of f(x).
$$\Rightarrow g'(f(x))\cdot f'(x)=1 \Rightarrow g'(f(x))=\frac{1}{f'(x)}$$

Also,
$$f'(x)=\frac{1}{3(x-8)^{2/3}}=\frac{1}{3f^2(x)}$$
Substituting this
$$g'(f(x))=3f^2(x) \Rightarrow g'(x)=3x^2$$

 

[HallsofIvy]

But in this particular case it is easier to find the inverse and then differentiate!

 

[Saitama]

But in this particular case it is easier to find the inverse and then differentiate!

Yes, it is. I just wanted to show an alternative approach because it won't be easy to find the inverse always. For example, I recently encountered the following problem:

Let $\displaystyle f(x)=\int_2^x \frac{dt}{\sqrt{1+t^4}}$ and g(x) be the inverse of f(x). Find g'(0).

Do you think it would be easy or even possible to find the inverse?

 

[jackmell]

Yes, it is. I just wanted to show an alternative approach because it won't be easy to find the inverse always. For example, I recently encountered the following problem:

Let $\displaystyle f(x)=\int_2^x \frac{dt}{\sqrt{1+t^4}}$ and g(x) be the inverse of f(x). Find g'(0).

Do you think it would be easy or even possible to find the inverse?

Looks like \sqrt{17} to me.

 

[ShreyasR]

$\displaystyle f(x)=\int_2^x \frac{dt}{\sqrt{1+t^4}}$

Could you find a solution to this integral?

 

[Saitama]

Could you find a solution to this integral?

I don't think so. Wolfram Alpha does not give a nice answer.

http://www.wolframalpha.com/input/?i=\int+\frac{dt}{\sqrt{1+t^4}}