# Thread: derivative of inverse function

1. ## 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...

2. 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 $f(x)$ has an inverse $f^{-1}(x)$, we have that $f(f^{-1}(x))=x$, for all $x$.
By the chain rule it follows from this that $f'(f^{-1}(x))\cdot (f^{-1})'(x)=1$, now divide both sides by $f'(f^{-1}(x))$.

3. Hello, bobey!

Prove that: . $\frac{d}{dx}(\arctan x) \:=\: \frac{1}{1+x^2}$
The usual method goes something like this . . .

Let: . $y \;=\;\arctan x$

Then: . $\tan y \:=\:x$ .[1]

Differentiate implicitly:

. . $\sec^2y\,\frac{dy}{dx} \:=\: 1 \quad\Rightarrow\quad \frac{dy}{dx} \;=\;\cos^2\!y$ .[2]

From [1], we have: . $\tan y \:=\:\frac{x}{1} \:=\:\frac{opp}{adj}$

Then $y$ is an angle in a right triangle with: $opp = x,\;adj = 1$
Pythagorus says: . $hyp \:=\:\sqrt{1+x^2}$
. . Hence: . $\cos y \;=\;\frac{adj}{hyp} \;=\;\frac{1}{\sqrt{1+x^2}}$

Therefore, [2] becomes: . $\frac{dy}{dx} \;=\;\left(\frac{1}{\sqrt{1+x^2}}\right)^2 \;=\;\frac{1}{1+x^2}$