What is the derivative of $f (x) = {tan}^{- 1} (x)$ $f(x)=tan^-1(x)$ ?

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What is the derivative of $f (x) = {tan}^{- 1} (x)$ $f(x)=tan^-1(x)$ ?

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Answer 1 · 2015-07-01T16:47:50

I seem to recall my professor forgetting how to deriving this. This is what I showed him:

$y = arctan x$ $y = arctanx$

$tan y = x$ $tany = x$

${sec}^{2} y \frac{d y}{d x} = 1$ $sec^2y (dy)/(dx) = 1$

$\frac{d y}{d x} = \frac{1}{{sec}^{2} y}$ $(dy)/(dx) = 1/(sec^2y)$

Since $tan y = \frac{x}{1}$ $tany = x/1$ and $\sqrt{1^{2} + x^{2}} = \sqrt{1 + x^{2}}$ $sqrt(1^2 + x^2) = sqrt(1+x^2)$ , ${sec}^{2} y = {(\frac{\sqrt{1 + x^{2}}}{1})}^{2} = 1 + x^{2}$ $sec^2y = (sqrt(1+x^2)/1)^2 = 1+x^2$

$\Rightarrow \frac{d y}{d x} = \frac{1}{1 + x^{2}}$ $=> color(blue)((dy)/(dx) = 1/(1+x^2))$

I think he originally intended to do this:

$\frac{d y}{d x} = \frac{1}{{sec}^{2} y}$ $(dy)/(dx) = 1/(sec^2y)$

${sec}^{2} y = 1 + {tan}^{2} y$ $sec^2y = 1+tan^2y$

${tan}^{2} y = x \to {sec}^{2} y = 1 + x^{2}$ $tan^2y = x -> sec^2y = 1+x^2$

$\Rightarrow \frac{d y}{d x} = \frac{1}{1 + x^{2}}$ $=> (dy)/(dx) = 1/(1+x^2)$

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What is the derivative of $f (x) = {tan}^{- 1} (x)$ $f(x)=tan^-1(x)$ ?

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