What is the derivative of ${tan}^{- 1} (4 x)$ $tan^-1(4x)$ ?

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What is the derivative of ${tan}^{- 1} (4 x)$ $tan^-1(4x)$ ?

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Answer 1 · 2015-04-23T13:31:03

$\frac{4}{16 x^{2} + 1}$ $4/(16x^2 +1)$

Derivative of ${tan}^{- 1} 4 x$ $tan^-1 4x$ can be written using the formula for the derivative of ${tan}^{- 1} x$ $tan^-1x$ = $\frac{1}{1 + x^{2}}$ $1/(1+x^2)$

$\frac{d}{d x} {tan}^{- 1} 4 x$ $d/dx tan^-1 4x$ = $\frac{1}{16 x^{2} + 1} \frac{d}{d x} (4 x)$ $1/(16x^2 +1) d/dx (4x)$

= $\frac{4}{16 x^{2} + 1}$ $4/(16x^2 +1)$

Answer 2 · 2015-04-23T19:58:25

$y = arctan (4 x)$ $y=arctan(4x)$

$tan y = 4 x$ $tany=4x$

${sec}^{2} y \cdot \frac{d y}{d x} = 4$ $sec^2y*(dy)/(dx)=4$

$({tan}^{2} y + 1) \cdot \frac{d y}{d x} = 4$ $(tan^2y+1)*(dy)/(dx)=4$

$(16 x^{2} + 1) \cdot \frac{d y}{d x} = 4$ $(16x^2+1)*(dy)/(dx)=4$

$\frac{d y}{d x} = \frac{4}{16 x^{2} + 1}$ $(dy)/(dx)=4/(16x^2+1)$

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What is the derivative of ${tan}^{- 1} (4 x)$ $tan^-1(4x)$ ?

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