If fI(x)=2x2−1 and y=f(x2) , then dydx atx=1 is

fI(x)=2x2−1 y=f(x2) dydx=fI(x2)2x dydx=2(x2)2−1×2x (dydx)x=1=2−1.2(1)=2

If fI(x)=2x2−1 and y=f(x2) , then dydx atx=1 is

fI(x)=2x2−1 y=f(x2) dydx=fI(x2)2x dydx=2(x2)2−1×2x (dydx)x=1=2−1.2(1)=2

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If $f^{I} (x) = \sqrt{2 x^{2} - 1}$ and $y = f (x^{2})$ , then $\frac{d y}{d x}$ at $x = 1$ is

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f^{I} (x) = \sqrt{2 x^{2} - 1}

$y = f (x^{2})$

$\frac{d y}{d x} = f^{I} (x^{2}) 2 x$

$\frac{d y}{d x} = \sqrt{2 {(x^{2})}^{2} - 1} \times 2 x$

${(\frac{d y}{d x})}_{x = 1} = \sqrt{2 - 1} .2 (1) = 2$

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If $f^{I} (x) = \sqrt{2 x^{2} - 1}$ and $y = f (x^{2})$ , then $\frac{d y}{d x}$ at $x = 1$ is