How do you find the derivative of ${tan}^{- 1} (x^{2})$ $tan^-1(x^2)$ ?

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Answer 1 · 2015-07-05T03:11:15

Assuming you don't remember the derivative of $arctan u$ $arctanu$ :

$y = {arctan x}^{2}$ $y = arctanx^2$

$tan y = x^{2}$ $tany = x^2$

${sec}^{2} y (\frac{d y}{d x}) = 2 x$ $sec^2y ((dy)/(dx)) = 2x$

$\frac{d y}{d x} = \frac{2 x}{{sec}^{2} y}$ $(dy)/(dx) = (2x)/sec^2y$

$= \frac{2 x}{1 + {tan}^{2} y}$ $= (2x)/(1+tan^2y)$

Since $tan y = x^{2}$ $tany = x^2$ :

$= \frac{2 x}{1 + x^{4}}$ $= color(blue)((2x)/(1+x^4)$

If you did remember it:

$\frac{d}{d x} [arctan u] = \frac{1}{1 + u^{2}} (\frac{d u}{d x})$ $d/(dx)[arctanu] = 1/(1+u^2)((du)/(dx))$

$u = x^{2}$ $u = x^2$

$\Rightarrow \frac{1}{1 + {(x^{2})}^{2}} \cdot 2 x = \frac{2 x}{1 + x^{4}}$ $=> 1/(1+(x^2)^2)*2x = color(darkblue)((2x)/(1+x^4))$

How do you find the derivative of tan^-1(x^2)tan−1(x2)tan^-1(x^2)?