- #1
UrbanXrisis
- 1,196
- 1
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
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