• October 22nd 2013, 01:07 PM
yanirose
Recall that f'(tanx) = sec^2(x) = tan^2(x) +1
By defining suitable f(x) and g(x) such that f(g(x)) = x and using the chain rule, find the derivative of tan^-1(x) [this is the same as arctan x ] with regard to x

have a nice day.
• October 22nd 2013, 01:19 PM
SlipEternal
What would you get for $\dfrac{d}{dx}f(g(x))$? (Hint: You can probably copy the answer directly from your book. I am just asking you to write the formula for the Chain Rule.)
• October 22nd 2013, 03:07 PM
HallsofIvy
"By defining suitable f(x) and g(x) such that f(g(x)) = x" means that f and g are inverse functions, right? So since you are given information about tan(x) and asked about $tan^{-1}(x)$. So what do you think "f" and "g" ought to be?
• October 22nd 2013, 04:26 PM
Prove It
\displaystyle \begin{align*} y &= \arctan{(x)} \\ \tan{(y)} &= x \\ \frac{d}{dx} \left[ \tan{(y)} \right] &= \frac{d}{dx} \left( x \right) \\ \sec^2{(y)}\,\frac{dy}{dx} &= 1 \\ \left[ 1 + \tan^2{(y)} \right] \frac{dy}{dx} &= 1 \\ \left( 1 + \left \{ \tan{ \left[ \arctan{(x)} \right] } \right\} ^2 \right) \frac{dy}{dx} &= 1 \\ \left( 1 + x^2 \right) \frac{dy}{dx} &=1 \\ \frac{dy}{dx} &= \frac{1}{1 + x^2} \end{align*}