यदि f(x) = (|x|)^{|sin x|} है,तो f'left( -fra | vedclass

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(A) दिया गया है $f(x) = |x|^{|\sin x|}$.दोनों पक्षों का प्राकृतिक लघुगणक लेने पर,$\ln f(x) = |\sin x| \ln |x|$.$x < 0$ के लिए,$|x| = -x$ और $|\sin x|

Vedclass · Accepted Answer

$(\frac{\pi}{4})^{1/\sqrt{2}} \left( \frac{\sqrt{2}}{2} \log \frac{4}{\pi} - \frac{2\sqrt{2}}{\pi} \right)$

Vedclass · Answer

(A) दिया गया है $f(x) = |x|^{|\sin x|}$.दोनों पक्षों का प्राकृतिक लघुगणक लेने पर,$\ln f(x) = |\sin x| \ln |x|$.$x अतः,$\ln f(x) = -\sin x \ln(-x)$.$x$ के सापेक्ष अवकलन करने पर:$\frac{f'(x)}{f(x)} = -\cos x \ln(-x) - \sin x \cdot \frac{1}{-x} \cdot (-1) = -\cos x \ln(-x) - \frac{\sin x}{x}$.$x = -\frac{\pi}{4}$ पर:$f(-\frac{\pi}{4}) = |-\frac{\pi}{4}|^{|\sin(-\frac{\pi}{4})|} = (\frac{\pi}{4})^{1/\sqrt{2}}$.$f'(-\frac{\pi}{4}) = f(-\frac{\pi}{4}) \left[ -\cos(-\frac{\pi}{4}) \ln(\frac{\pi}{4}) - \frac{\sin(-\frac{\pi}{4})}{-\frac{\pi}{4}} \right]$.$f'(-\frac{\pi}{4}) = (\frac{\pi}{4})^{1/\sqrt{2}} \left[ -\frac{1}{\sqrt{2}} \ln(\frac{\pi}{4}) - \frac{-1/\sqrt{2}}{\pi/4} \right]$.$f'(-\frac{\pi}{4}) = (\frac{\pi}{4})^{1/\sqrt{2}} \left[ \frac{1}{\sqrt{2}} \ln(\frac{4}{\pi}) - \frac{4}{\sqrt{2}\pi} \right]$.$f'(-\frac{\pi}{4}) = (\frac{\pi}{4})^{1/\sqrt{2}} \left[ \frac{\sqrt{2}}{2} \ln(\frac{4}{\pi}) - \frac{2\sqrt{2}}{\pi} \right]$.

यदि $f(x) = (|x|)^{|\sin x|}$ है,तो $f'\left( -\frac{\pi}{4} \right) = $

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