જો f(x) = int_{pi^2/16}^{x^2} frac{sin x cdot | vedclass

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(A) આપેલ વિધેય: $f(x) = \sin x \int_{\pi^2/16}^{x^2} \frac{\sin \sqrt{	heta}}{1 + \cos^2 \sqrt{	heta}} \, d	heta$. લીબનીઝના નિયમનો ઉપયોગ કરીને વિકલ

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$\pi$

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(A) આપેલ વિધેય: $f(x) = \sin x \int_{\pi^2/16}^{x^2} \frac{\sin \sqrt{	heta}}{1 + \cos^2 \sqrt{	heta}} \, d	heta$. લીબનીઝના નિયમનો ઉપયોગ કરીને વિકલન કરતા: $f'(x) = \cos x \int_{\pi^2/16}^{x^2} \frac{\sin \sqrt{	heta}}{1 + \cos^2 \sqrt{	heta}} \, d	heta + \sin x \left[ \frac{\sin \sqrt{x^2}}{1 + \cos^2 \sqrt{x^2}} \cdot \frac{d}{dx}(x^2) ight]$. $f'(x) = \cos x \int_{\pi^2/16}^{x^2} \frac{\sin \sqrt{	heta}}{1 + \cos^2 \sqrt{	heta}} \, d	heta + \sin x \left[ \frac{\sin x}{1 + \cos^2 x} \cdot 2x ight]$. હવે,$x = \frac{\pi}{2}$ મૂકતા: $f'(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) \int_{\pi^2/16}^{\pi^2/4} \frac{\sin \sqrt{	heta}}{1 + \cos^2 \sqrt{	heta}} \, d	heta + \sin(\frac{\pi}{2}) \left[ \frac{\sin(\frac{\pi}{2})}{1 + \cos^2(\frac{\pi}{2})} \cdot 2(\frac{\pi}{2}) ight]$. કારણ કે $\cos(\frac{\pi}{2}) = 0$ અને $\sin(\frac{\pi}{2}) = 1$: $f'(\frac{\pi}{2}) = 0 \cdot (	ext{સંકલિતનું મૂલ્ય}) + 1 \cdot \left[ \frac{1}{1 + 0} \cdot \pi ight] = \pi$.

જો $f(x) = \int_{\pi^2/16}^{x^2} \frac{\sin x \cdot \sin \sqrt{\theta}}{1 + \cos^2 \sqrt{\theta}} \, d\theta$ હોય,તો $f'(\frac{\pi}{2})$ નું મૂલ્ય શોધો.

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