Practicing Success
If $f(x)=|x|^{|\sin x|}$, then $f'\left(-\frac{\pi}{4}\right)$ equals |
$\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \ln \frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}\right)$ $\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \ln \frac{4}{\pi}+\frac{2 \sqrt{2}}{\pi}\right)$ $\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \ln \frac{\pi}{4}-\frac{2 \sqrt{2}}{\pi}\right)$ $\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \ln \frac{\pi}{4}+\frac{2 \sqrt{2}}{\pi}\right)$ |
$\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \ln \frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}\right)$ |
In the neighbourhood of $-\pi / 4$, we have $f(x) =(-x)^{-\sin x}=e^{-\sin x \log (-x)}$ $\Rightarrow f'(x) =e^{-\sin x \log (-x)}\left(-\cos x . \log (-x)-\frac{\sin x}{x}\right)$ $\Rightarrow f'(x)=(-x)^{-\sin x}\left(-\cos x . \log (-x)-\frac{\sin x}{x}\right)$ $\Rightarrow f'\left(-\frac{\pi}{4}\right)=\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{-1}{\sqrt{2}} \log \frac{\pi}{4}+\frac{4}{\pi} \times \frac{-1}{\sqrt{2}}\right)$ $\Rightarrow f'\left(-\frac{\pi}{4}\right)=\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \log \frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}\right)$ |