Question

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

• A. 2
• B. 1
• C. -2
• D. -1

Answer 1

The correct option is A 2

$f^{\prime }\left(x\right)=\sqrt{2x^2-1}$ $y=f\left(x^2\right)$
$\begin{array}{c}\frac{dy}{dx}=f^{\prime }\left(x^2\right)\times \left[2x\right]\\ now{f}^{\prime }\left(x\right)=\sqrt{2x^2-1}\\ f^{\prime }\left(x^2\right)=\sqrt{2x^4-1}\\ \therefore \frac{dy}{dx}=\sqrt{2x^4-1}\times 2x=2\end{array}$