यदि f(x) = frac{sin^{2} x}{1+cot x} + frac{co | vedclass

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(C) दिया गया है,$f(x) = \frac{\sin^{2} x}{1+\cot x} + \frac{\cos^{2} x}{1+	an x}$.व्यंजक को सरल करने पर:$f(x) = \frac{\sin^{2} x}{1+\frac{\cos x}{\si

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(C) दिया गया है,$f(x) = \frac{\sin^{2} x}{1+\cot x} + \frac{\cos^{2} x}{1+	an x}$.व्यंजक को सरल करने पर:$f(x) = \frac{\sin^{2} x}{1+\frac{\cos x}{\sin x}} + \frac{\cos^{2} x}{1+\frac{\sin x}{\cos x}}$$f(x) = \frac{\sin^{3} x}{\sin x + \cos x} + \frac{\cos^{3} x}{\cos x + \sin x}$$f(x) = \frac{\sin^{3} x + \cos^{3} x}{\sin x + \cos x}$सर्वसमिका $a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2})$ का उपयोग करने पर:$f(x) = \frac{(\sin x + \cos x)(\sin^{2} x - \sin x \cos x + \cos^{2} x)}{\sin x + \cos x}$$f(x) = \sin^{2} x - \sin x \cos x + \cos^{2} x$$f(x) = 1 - \sin x \cos x = 1 - \frac{1}{2} \sin(2x)$.अब,$x$ के सापेक्ष अवकलन करने पर:$f^{\prime}(x) = -\frac{1}{2} \cdot \cos(2x) \cdot 2 = -\cos(2x)$.$x = \frac{\pi}{4}$ रखने पर:$f^{\prime}\left(\frac{\pi}{4}ight) = -\cos\left(2 \cdot \frac{\pi}{4}ight) = -\cos\left(\frac{\pi}{2}ight) = 0$.

यदि $f(x) = \frac{\sin^{2} x}{1+\cot x} + \frac{\cos^{2} x}{1+\tan x}$ है,तो $f^{\prime}\left(\frac{\pi}{4}\right)$ है

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