Finding the Derivative of an Inverse Function

[UrbanXrisis]

I was taught in class that if g(x) is f^-1(x), then g'(y)=1/f'(x)

my notes for an example:
f(x)=x^2
g(x)=f^-1(x)=x^(.5)
g’(4)=?
g(x)= x^(.5)
g’(x)=.5x^-.5
g’(4)=1/f’(x), x=4^(.5)
g’(4)=1/f'(2)=1/2*2=1/4

I have no clue how g(x)=f^-1(x)=x^(.5) and how g’(4)=1/f'(2)

I am just overall confused any help would be appreciated

 

[Zlex]

f(x) = x^2

set f(x) = y

y = x^2

(*To find inverse solve for x explicitly in terms of y*)

y^0.5 = x

f^-1(x) = x^0.5

let g(x) = f^-1(x)

g(x) = x^0.5

The second one follows the same way

 

[UrbanXrisis]

I have:
g’(4)=1/f'(2)=1/2*2=1/4

and also:
g'(4)=1/f'(x)=1/f'(2)=1/4

is both ways acceptable?

 

[Zlex]

I have:
g’(4)=1/f'(2)=1/2*2=1/4

and also:
g'(4)=1/f'(x)=1/f'(2)=1/4

is both ways acceptable?

I wouldn't go with the second one. It could be seen as wrong to say g'(4) = 1/f'(x) without saying what x is explicitly, which in this case is root(4).

But, it depends how picky your instructor is

 

[UrbanXrisis]

what about:
g’(x)=.5x^-.5=.5(4)^-.5=1/4

could that be acceptable too?

 

[Zlex]

what about:
g’(x)=.5x^-.5=.5(4)^-.5=1/4

could that be acceptable too?

Sure,

g(x) = $$x^\frac{1}{2}$$

thus,

g'(x) = $$\frac{1}{2x^\frac{1}{2}}$$

g'(4) = $$\frac{1}{2*4^\frac{1}{2}}$$

g'(4) = $$\frac{1}{2*2}$$

g'(4) = $$\frac{1}{4}$$


Again, I would just recommend that you specify you are changing from the general equation g'(x) to g'(4) somewhere in your solution. But if its not on a test, or your teacher isn't picky then that should be fine.