Thread: derivative of inverse function

prove that d/dx arctan x = 1/(1+x^2)

the solution given by the lecturer...

(f-1)'(x) = 1/f'(f-1(x)) ....... by theorem 1
= 1/cos(arcsin x)
= 1/sqrt(1-sin^2(arcsin x)
= 1/sqrt(1-x^2)



just wandering how theorem 1 is derived.... =p anyone can explain to me... need to understand why rather than memorizing the formula...

Originally Posted by bobey

prove that d/dx arctan x = 1/(1+x^2)

the solution given by the lecturer...

(f-1)'(x) = 1/f'(f-1(x)) ....... by theorem 1
= 1/cos(arcsin x)
= 1/sqrt(1-sin^2(arcsin x)
= 1/sqrt(1-x^2)



just wandering how theorem 1 is derived.... =p anyone can explain to me... need to understand why rather than memorizing the formula...

Assuming that has an inverse , we have that , for all .
By the chain rule it follows from this that , now divide both sides by .

Hello, bobey!

Prove that: .

The usual method goes something like this . . .


Let: .

Then: . .[1]


Differentiate implicitly:

. . .[2]


From [1], we have: .

Then is an angle in a right triangle with:
Pythagorus says: .
. . Hence: .


Therefore, [2] becomes: .