Question

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

• A. $1$
• B. $-1$
• C. $-2$
• D. $2$

Answer 1

The correct option is D $2$

$y=\sqrt{2x^4-1}$

$\frac{dy}{dx}=\frac{1}{2}(2x^4-1)\times 4x$

$=2x(2x^4-1)$

${\frac{dy}{dx}}_{x=1}=2\left(1\right)\left(2\right(1{)}^4-1)$

$=2\left(1\right)$

$=2$.