If varphi (x) = int_{1/x}^{sqrt{x}} sin(t^2) | vedclass

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(C) Using the Leibniz integral rule,if $\varphi(x) = \int_{g(x)}^{h(x)} f(t) \, dt$,then $\varphi'(x) = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)$.Her

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$\frac{3}{2} \sin 1$

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(C) Using the Leibniz integral rule,if $\varphi(x) = \int_{g(x)}^{h(x)} f(t) \, dt$,then $\varphi'(x) = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)$.Here,$f(t) = \sin(t^2)$,$h(x) = \sqrt{x}$,and $g(x) = \frac{1}{x}$.Calculating the derivatives: $h'(x) = \frac{1}{2\sqrt{x}}$ and $g'(x) = -\frac{1}{x^2}$.Substituting these into the formula:$\varphi'(x) = \sin((\sqrt{x})^2) \cdot \frac{1}{2\sqrt{x}} - \sin((\frac{1}{x})^2) \cdot (-\frac{1}{x^2})$$\varphi'(x) = \sin(x) \cdot \frac{1}{2\sqrt{x}} + \frac{1}{x^2} \sin(\frac{1}{x^2})$.Now,evaluate at $x = 1$:$\varphi'(1) = \sin(1) \cdot \frac{1}{2\sqrt{1}} + \frac{1}{1^2} \sin(\frac{1}{1^2})$$\varphi'(1) = \frac{1}{2} \sin(1) + \sin(1) = \frac{3}{2} \sin(1)$.

If $\varphi (x) = \int_{1/x}^{\sqrt{x}} \sin(t^2) \, dt$,then $\varphi'(1) = $

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