I was having trouble with the following question about finding the derivative of:

y=(tan^-1(x/2))^4

-i utilized the chain rule but this came out to be wrong.

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- Jun 4th 2011, 09:20 PMjohnsy123Derivative of inverse tan fuction
I was having trouble with the following question about finding the derivative of:

y=(tan^-1(x/2))^4

-i utilized the chain rule but this came out to be wrong. - Jun 4th 2011, 09:23 PMChris L T521
- Jun 4th 2011, 09:42 PMProve It
$\displaystyle \displaystyle \begin{align*} y &= \left[\arctan{\left(\frac{x}{2}\right)}\right]^4 \\ y^{\frac{1}{4}} &= \arctan{\left(\frac{x}{2}\right)} \\ \tan{\left(y^{\frac{1}{4}}\right)} &= \frac{x}{2} \\ \frac{d}{dx}\left[\tan{\left(y^{\frac{1}{4}}\right)}\right] &= \frac{d}{dx}\left(\frac{x}{2}\right) \\ \frac{d}{dy}\left[\tan{\left(y^{\frac{1}{4}}\right)}\right]\,\frac{dy}{dx} &= \frac{1}{2} \\ \frac{1}{4}y^{-\frac{3}{4}}\sec^2{\left(y^{\frac{1}{4}}\right)}\, \frac{dy}{dx} &= \frac{1}{2} \\ \frac{dy}{dx} &= 2y^{\frac{3}{4}}\cos^2{\left(y^{\frac{1}{4}}\right )}\end{align*}$

You could convert it back to a function of $\displaystyle \displaystyle x$ now if you want.