What is the derivative of inverse tangent of 2x?

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Answer 1 · 2015-04-14T14:38:10

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

Solution
Let

$y = {tan}^{- 1} 2 x$ $y=tan^(-1)2x$

$tan y = 2 x$ $tany=2x$

Differentiating both side with respect to 'x'

$\frac{d}{d x} (tan y) = \frac{d}{d x} (2 x)$ $d/dx(tany)=d/dx(2x)$

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

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

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

Now, as
$tan y = 2 x$ $tany=2x$

${tan}^{2} y = {(2 x)}^{2}$ $tan^2y=(2x)^2$

${tan}^{2} y = 4 x^{2}$ $tan^2y=4x^2$

So,

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

Answer 2 · 2015-04-14T14:44:41

dy/(dx)=2/(1+4x^2)#

Solution

Let

$y = {tan}^{- 1} 2 x$ $y=tan^(-1)2x$

Differentiating both side with respect to 'x'

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

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

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

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

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

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