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. 2
• B. 1
• C. -2
• D. none of these

Answer 1

Answer: (D)

none of these

We have,

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

And

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

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

$y=2x^2-1$

On differentiating and we get,

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

$\frac{dy}{dx}=4x-0$

$\frac{dy}{dx}=4x$

At point $(x=1)$

So,

$\frac{dy}{dx}=4x=4\left(1\right)$

$\frac{dy}{dx}=4$

Hence, this is the answer.