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Question: If y =\(\int_{0}^{x}\frac{t^{2}}{\sqrt{t^{2} + 1}}\)dt then the rate of change of y w.r.t. x when x...

If y =0xt2t2+1\int_{0}^{x}\frac{t^{2}}{\sqrt{t^{2} + 1}}dt then the rate of change of y w.r.t. x when

x = 1 is

A

12\frac{1}{2}

B

2\sqrt{2}

C

12\frac{1}{\sqrt{2}}

D

None of these

Answer

12\frac{1}{\sqrt{2}}

Explanation

Solution

dydx=x2x2+1.1\frac{dy}{dx} = \frac{x^{2}}{\sqrt{x^{2} + 1}}.1; (dydx)atx=1=12\left( \frac{dy}{dx} \right)_{atx = 1} = \frac{1}{\sqrt{2}}