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# What is the derivative of f(x)=tan^-1(x) ?

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#### Explanation

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#### Explanation:

I want someone to double check my answer

18
Jul 1, 2015

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

$y = \arctan x$

$\tan y = x$

${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{\sec}^{2} y}$

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

$\implies \textcolor{b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {x}^{2}}}$

I think he originally intended to do this:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{\sec}^{2} y}$

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

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

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {x}^{2}}$

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