# Derivative of inverse function

Staff member

#### Dr.Peterson

##### Elite Member
If f(x) = 2x - 1/x for x>0 and g is the inverse of f, then what is the value of g’(0)
It looks like you are finding the value of g(0), that is, x such that f(x) = 0. That's good.

What have you learned about the derivative of an inverse function? What have you done on that part of the problem?

#### pka

##### Elite Member
$$\displaystyle f(x)=2x-\frac{1}{x},~x>0$$ Suppose the $$\displaystyle g$$ is the inverse of $$\displaystyle f$$ on its domain.
Here is the idea. Use implicit differentiation:
$$\displaystyle g(f(x))=x\\g'(f(x))\cdot f'(x)=1\\g'(f(x))=\dfrac{1}{f'(x)}$$
Now you need $$\displaystyle f(?)=0$$

#### Anbinh13

##### New member
If f(x) = 2x - 1/x for x>0 and g is the inverse of f, then what is the value of g’(0)View attachment 15139
I just need the correct answer because the answer I got what different compare to my teacher.

#### Subhotosh Khan

##### Super Moderator
Staff member
"I just need the correct answer because the answer I got what different compare to my teacher."
What answer did you get? More importantly,

HOW did you get it?

Show us - so that we can show you the correct way!

#### pka

##### Elite Member
@Anbinh13, It was completely rude of you to quote your post back to us.
We can read. But we wanted you to provide some of your work so that we can see if you need help understanding the concepts incolved, Prof Peterson asked you about your knowledge of the inverse function. You completely ignored his request. WHY?

#### Jomo

##### Elite Member
@Anbinh13, I agree with pka that it was rude of you to reply with your prior post. It is even more rude for you to just ask us to provide you with the answer. This is a help site, where we help you. A private tutor is what you want.

#### HallsofIvy

##### Elite Member
As was pointed out before to find, first, g(0), you need to solve the equation, f(x)= 2x- 1/x= 0. You should see that multiplying both sides by x gives a quadratic that has two roots. Be sure to check each back in the original equation.