Question

Derivative of sin^-1(2x/1+x^2)

Answer 1

Hey there!

Answer:

Let y = $sin^{-1} \dfrac{2x}{1 + x^{2}}$


Now put x = tan$\theta$


Therefore, $\theta = tan^{-1} x$


y = $\sin^{-1} ( \dfrac{2tan\theta}{1 + tan^{2} \theta} )$


We know 2tan$\theta$/ 1 + $tan^{2}$ = sin2$\theta$


Using the above formula:


y = $sin^{-1} ( sin2\theta )$


y = $\cancel{sin^{-1}} (\cancel{sin}2\theta )$


y = 2$\theta$


Substituting value of $\theta$


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


Now differentiate both sides wrt x:


$\boxed{\dfrac{dy}{dx} = 2 \dfrac{1}{1 + x^{2}}}$



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