Let f(x) = int_{sin x}^{cos x} e^{-t^2} dt. T | vedclass

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(B) Using the Leibniz integral rule, the derivative of $f(x) = \int_{g(x)}^{h(x)} F(t) dt$ is given by $f^{\prime}(x) = F(h(x)) \cdot h^{\prime}(x) -

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$-\sqrt{2/e}$

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(B) Using the Leibniz integral rule, the derivative of $f(x) = \int_{g(x)}^{h(x)} F(t) dt$ is given by $f^{\prime}(x) = F(h(x)) \cdot h^{\prime}(x) - F(g(x)) \cdot g^{\prime}(x)$.Here, $F(t) = e^{-t^2}$, $h(x) = \cos x$, and $g(x) = \sin x$.So, $f^{\prime}(x) = e^{-(\cos x)^2} \cdot (-\sin x) - e^{-(\sin x)^2} \cdot (\cos x)$.At $x = \frac{\pi}{4}$, we have $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.Substituting these values:$f^{\prime}\left(\frac{\pi}{4}\right) = -e^{-1/2} \cdot \frac{1}{\sqrt{2}} - e^{-1/2} \cdot \frac{1}{\sqrt{2}}$$f^{\prime}\left(\frac{\pi}{4}\right) = -\frac{2}{\sqrt{2} \cdot \sqrt{e}} = -\frac{\sqrt{2}}{\sqrt{e}} = -\sqrt{\frac{2}{e}}$.

Let $f(x) = \int_{\sin x}^{\cos x} e^{-t^2} dt$. Then $f^{\prime}\left(\frac{\pi}{4}\right)$ equals

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