If f-x-2x2-1 and y-f x2 then dy-dx at x-1 is

The correct option is A 2 Given, #160; f^'(x)=√(2x^2 1) y=f ( x^2 ) dydx=f'(x^2)2x #160; #160; #160; #160; #160; #160; ...

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If f'x=√2x2...

Question

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

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Similar questions

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

y = f (x^{2})

\frac{d y}{d x}

x = 1

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

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

\frac{d y}{d x}

x = 1

f^{'} (x) = sin x^{2}

y = f (x^{2} + 1)

\frac{d y}{d x}

x = 2

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

y = f (x^{2})

\frac{d y}{d x}

x = 1

f' (x) = \sqrt{2 x^{2} - 1} and y = f (x^{2})

\frac{d y}{d x} a t x = 1

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