Question

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

• A. $4$
• B. $1$
• C. $-2$
• D. $-1$

Answer 1

Answer: (A)

$4$

$f\left(x\right)=\sqrt{2x^2-1}$

$y=f\left(x^2\right)$

$y^1{|}_{x=1}\Rightarrow \frac{d}{dx}\sqrt{2(x^2{)}^2-1}$

$y^1=\frac{1}{2}\frac{2.4x^3}{2.x^4-1}{|}_1$

$\Rightarrow \frac{1}{2}\frac{2.4(1{)}^3}{2\sqrt{2(1{)}^4-1}}$

$y^1\Rightarrow \frac{1}{2}\frac{2.4}{\sqrt{2-1}}=\frac{1}{2}.\frac{8}{1}\Rightarrow 4$