# Derivatives of Inverse Functions

• November 12th 2013, 02:50 PM
nycmath
Derivatives of Inverse Functions
See picture.
• November 12th 2013, 07:04 PM
SlipEternal
Re: Derivatives of Inverse Functions
If $\sin\theta = \dfrac{t}{1} = \dfrac{\text{opp}}{\text{hyp}}$, then $\theta = \sin^{-1}(t)$.

So $\tan(\sin^{-1}(t)) = \dfrac{\text{opp}}{\text{adj}} = \dfrac{t}{\sqrt{1-t^2}}$.

Hence $g(t) = \dfrac{t}{\sqrt{1-t^2}}$.

So, after applying the quotient rule, multiply top and bottom by $\sqrt{1-t^2}$:

\begin{align*}g'(t) & = \dfrac{\sqrt{1-t^2} + \dfrac{t^2}{\sqrt{1-t^2}}}{1-t^2}\cdot \dfrac{\sqrt{1-t^2}}{\sqrt{1-t^2}} \\ & = \dfrac{1-t^2+t^2}{(1-t^2)^{3/2}} \\ & = \dfrac{1}{(1-t^2)^{3/2}}\end{align*}
• November 12th 2013, 07:15 PM
Prove It
Re: Derivatives of Inverse Functions
Quote:

Originally Posted by nycmath
See picture.

If you're going to attempt using the chain rule, let \displaystyle \begin{align*} u = \arcsin{(t)} \implies y = \tan{(u)} \end{align*}. Then \displaystyle \begin{align*} \frac{du}{dt} = \frac{1}{\sqrt{1 - t^2}} \end{align*} and

\displaystyle \begin{align*} \frac{dy}{du} &= \sec^2{(u)} \\ &= \sec^2{\left[ \arcsin{(t)} \right] } \\ &= \frac{1}{\cos^2{ \left[ \arcsin{(t)} \right] } } \\ &= \frac{1}{1 - \sin^2{ \left[ \arcsin{(t)} \right] } } \\ &= \frac{1}{1 - t^2} \end{align*}

Thus

\displaystyle \begin{align*} \frac{dy}{dt} &= \frac{du}{dt} \cdot \frac{dy}{du} \\ &= \frac{1}{\sqrt{ 1 - t^2} } \cdot \frac{1}{1 - t^2} \\ &= \frac{1}{ \left( 1 - t^2 \right) ^{\frac{3}{2}} } \end{align*}
• November 13th 2013, 05:29 AM
nycmath
Re: Derivatives of Inverse Functions
I now know what I did wrong. Than you very much.