Question

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

Answer 1

We have,
$y=f\left(x^2\right)\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=f\text{'}\left(x^2\right)\cdot 2x\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2x\sqrt{2x^4-1}\therefore {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{x=1}=2\left(1\right)\sqrt{2{\left(1\right)}^4-1}=2$