What is the derivative of inverse tangent of 2x?

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Answer 1 · 2015-04-14T14:38:10

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

Solution
Let

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

$tany=2x$

Differentiating both side with respect to 'x'

$d/dx(tany)=d/dx(2x)$

$=>sec^2y(dy/(dx))=2$

$=>dy/(dx)=2/(sec^2y)$

$=>dy/(dx)=2/(1+tan^2y)$

Now, as
$tany=2x$

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

$tan^2y=4x^2$

So,

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

Answer 2 · 2015-04-14T14:44:41

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

Solution

Let

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

Differentiating both side with respect to 'x'

$dy/dx=d/dx(tan^(-1)2x)$

$dy/dx=1/(1+(2x)^2)d/dx(2x)$

$dy/dx=1/(1+(2x)^2).2d/dx(x)$

$dy/dx=1/(1+(2x)^2).2dx/dx$

$dy/dx=1/(1+(2x)^2).2$

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