Find the derivative of the $y=\tan^{-1}(\frac{3x-x^3}{1-3x^3})$
Answer & explanation
Correct answer: option 4
Put $x=\tan \theta$. Then $y=\tan^{-1}(\tan 3\theta)=3\theta=3\tan^{-1}x$. Differentiating w.r.to x we get $\frac{dy}{dx}=\frac{3}{1+x^2}$
Find the derivative of the $y=\tan^{-1}(\frac{3x-x^3}{1-3x^3})$
Correct answer: option 4
Put $x=\tan \theta$. Then $y=\tan^{-1}(\tan 3\theta)=3\theta=3\tan^{-1}x$. Differentiating w.r.to x we get $\frac{dy}{dx}=\frac{3}{1+x^2}$