y=∫xx25−t2dt then the value of dydx at x=2 is

y=∫xx25−t2dtdydx=ddx(∫xx25−t2dt) [by newton's Leibnity Rule]=5−(x2)2d(x2)−5−x2d(x)=5−x4(2x)−5−x2At x=2dydx=5−4(22)−5−2 =22−3

y=∫xx25−t2dt then the value of dydx at x=2 is

y=∫xx25−t2dtdydx=ddx(∫xx25−t2dt) [by newton's Leibnity Rule]=5−(x2)2d(x2)−5−x2d(x)=5−x4(2x)−5−x2At x=2dydx=5−4(22)−5−2 =22−3

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$y =$ $\int_{x}^{x^{2}} \sqrt{5 - t^{2}} d t$ then the value of $\frac{d y}{d x}$ at $x = \sqrt{2}$ is

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$1 - \sqrt{3}$

$\sqrt{3} (2 \sqrt{6} - 1)$

$2 \sqrt{2} - \sqrt{3}$

$2 \sqrt{2} + \sqrt{3}$

answer is C.

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Detailed Solution

$\begin{array}{l} y = \int_{x}^{x^{2}} \sqrt{5 - t^{2}} d t \\ \frac{d y}{d x} = \frac{d}{d x} (\int_{x}^{x^{2}} \sqrt{5 - t^{2}} d t) [b y n e w t o n' s L e i b n i t y R u l e] \\ = \sqrt{5 - {(x^{2})}^{2}} d (x^{2}) - \sqrt{5 - x^{2}} d (x) \\ = \sqrt{5 - x^{4}} (2 x) - \sqrt{5 - x^{2}} \\ A t x = \sqrt{2} \\ \frac{d y}{d x} = \sqrt{5 - 4} (2 \sqrt{2}) - \sqrt{5 - 2} \\ = 2 \sqrt{2} - \sqrt{3} \end{array}$