Question

If $f^{\prime }\left(x\right)=\sqrt{2x^2-1}$ and $y=f\left(x^2\right)$ then find $\frac{dy}{dx}$ at $x=1$

Answer 1

Given $y=f\left(x^2\right)$
$\frac{dy}{dx}=f^{\prime }\left(x^2\right)\times 2x=2x\sqrt{2x^4-1}$
So at $x=1$
$\frac{dy}{dx}=2\left(1\right)\sqrt{2(1{)}^4-1}=2$