Question:
If $f^{\prime}(x)=\sqrt{2 x^{2}-1}$ and $y=f\left(x^{2}\right)$, then find at $x=1$.
Solution:
$y=f\left(x^{2}\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}\left(\mathrm{x}^{2}\right) \cdot 2 \mathrm{x}$
$=2 x \sqrt{2\left(x^{2}\right)^{2}-1}$
$=2 x \sqrt{2 x^{4}-1}$
Putting $x=1$,
$\frac{d y}{d x}=2 \cdot 1 \cdot \sqrt{2 \cdot 1^{4}-1}$
$=2 \sqrt{2-1}$
$=2 \sqrt{1}$
$=2$
i.e., $\frac{\mathrm{dy}}{\mathrm{dx}}=2$ at $\mathrm{x}=1$. (Ans)