Question

differentiate
$\sin2 x \: \: with \: { \tan }^{ - 1}$
how to solve it​

Answer 1

I am assuming you are asking for derivative of sin(2x) with respect to tan^(-1)(x).

If f and g are two differentiable functions in variable 'x', then

$\frac{df}{dg} = \frac{ \frac{df}{dx} }{ \frac{dg}{dx} }$

So,

$\frac{d(sin(2x))}{d(tan^{ - 1}(x)) } = \frac{ \frac{d(sin(2x))}{dx} }{ \frac{d(tan^{ - 1}(x)) }{dx} } \\ = \frac{2cos(2x)}{ \frac{1}{1 + x^{2} } } \\ = 2(1 + x^{2})(cos(2x))$

Hope, this helps...