Question

$y = tan^{-1} (\frac{4 sin 2x}{cos 2x - 6sin^2x})$ then $\frac{dy}{dx}$ at $x=0$.

Answer 1

To find $\frac{dx}{dt}$ of $tan^{-1}[\frac{(cos(2x)-6sin^2(x))}{(4sin^2(x))}]$ at $x=0$,

we first substitute $u=\frac{(cos(2x)-6sin^2(x))}{(4sin^2(x))}$.

Then, using the chain rule, we get $\frac{dx}{dt}$ of $tan^{-1}(u)$ times $\frac{dx}{dt}$ of u.

Simplifying the expression and evaluating it at $x=0$, we get the final answer as 8.