If f'(x)= 2x^2-1 and y=f(x^2), then find dy dx at x=1. - Vidyadip

Question

If $\text{f}'\text{(x)}=\sqrt{2\text{x}^2-1}$ and $\text{y}=\text{f}(\text{x}^2),$ then find $\frac{\text{dy}}{\text{dx}}\text{at x}=1.$

✓

Answer

Here,
$\text{f}'\text{(x)}=\sqrt{2\text{x}^2-1}$
and $\text{y}=\text{f}\big(\text{x}^2\big)$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\text{f}\big(\text{x}^2\big)$
$=\text{f}'\big(\text{x}^2\big)\frac{\text{d}}{\text{dx}}\big(\text{x}^2\big)$
$=\text{d}'\big(\text{x}^2\big)\times2\text{x}$
$\frac{\text{dy}}{\text{dx}}=2\text{xf}'\big(\text{x}^2\big)$
Put x = 1
$\frac{\text{dy}}{\text{dx}}=2(1)\text{f}'(1)$
$=2\times\text{f}'(1)$
$\frac{\text{dy}}{\text{dx}}=2\times1$
$\big[\text{Since},\text{f}'(1)=\sqrt{2(1)^2-1}=\sqrt{2-1}=1\big]$
$\frac{\text{dy}}{\text{dx}}=2$

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