If f(x) = (|x|)^{|sin x|},then f'left( -frac{ | vedclass

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(A) Given $f(x) = |x|^{|\sin x|}$.Taking natural logarithm on both sides,$\ln f(x) = |\sin x| \ln |x|$.For $x < 0$,$|x| = -x$ and $|\sin x| = -\sin x$

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

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(A) Given $f(x) = |x|^{|\sin x|}$.Taking natural logarithm on both sides,$\ln f(x) = |\sin x| \ln |x|$.For $x So,$\ln f(x) = -\sin x \ln(-x)$.Differentiating with respect to $x$:$\frac{f'(x)}{f(x)} = -\cos x \ln(-x) - \sin x \cdot \frac{1}{-x} \cdot (-1) = -\cos x \ln(-x) - \frac{\sin x}{x}$.At $x = -\frac{\pi}{4}$:$f(-\frac{\pi}{4}) = |-\frac{\pi}{4}|^{|\sin(-\frac{\pi}{4})|} = (\frac{\pi}{4})^{1/\sqrt{2}}$.$f'(-\frac{\pi}{4}) = f(-\frac{\pi}{4}) \left[ -\cos(-\frac{\pi}{4}) \ln(\frac{\pi}{4}) - \frac{\sin(-\frac{\pi}{4})}{-\frac{\pi}{4}} \right]$.$f'(-\frac{\pi}{4}) = (\frac{\pi}{4})^{1/\sqrt{2}} \left[ -\frac{1}{\sqrt{2}} \ln(\frac{\pi}{4}) - \frac{-1/\sqrt{2}}{\pi/4} \right]$.$f'(-\frac{\pi}{4}) = (\frac{\pi}{4})^{1/\sqrt{2}} \left[ \frac{1}{\sqrt{2}} \ln(\frac{4}{\pi}) - \frac{4}{\sqrt{2}\pi} \right]$.$f'(-\frac{\pi}{4}) = (\frac{\pi}{4})^{1/\sqrt{2}} \left[ \frac{\sqrt{2}}{2} \ln(\frac{4}{\pi}) - \frac{2\sqrt{2}}{\pi} \right]$.

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

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