# Is #f(x)=e^-xcos(-x)-sinx/(pi-e^x)# increasing or decreasing at #x=pi/6#?

Given function:

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To determine whether ( f(x) = e^{-x}\cos(-x) - \frac{\sin x}{\pi - e^x} ) is increasing or decreasing at ( x = \frac{\pi}{6} ), we need to find the sign of its derivative at that point.

Calculate the derivative of ( f(x) ) with respect to ( x ):

[ f'(x) = -e^{-x}\cos(-x) + e^{-x}\sin(-x) - \frac{-e^x\sin x - e^x\cos x}{(\pi - e^x)^2} ]

Evaluate ( f'(x) ) at ( x = \frac{\pi}{6} ):

[ f'\left(\frac{\pi}{6}\right) = -e^{-\frac{\pi}{6}}\cos\left(-\frac{\pi}{6}\right) + e^{-\frac{\pi}{6}}\sin\left(-\frac{\pi}{6}\right) - \frac{-e^{\frac{\pi}{6}}\sin\frac{\pi}{6} - e^{\frac{\pi}{6}}\cos\frac{\pi}{6}}{(\pi - e^{\frac{\pi}{6}})^2} ]

[ = -e^{-\frac{\pi}{6}}\cos\left(\frac{\pi}{6}\right) + e^{-\frac{\pi}{6}}\sin\left(\frac{\pi}{6}\right) - \frac{-e^{\frac{\pi}{6}}\sin\frac{\pi}{6} - e^{\frac{\pi}{6}}\cos\frac{\pi}{6}}{\left(\pi - e^{\frac{\pi}{6}}\right)^2} ]

[ \approx -0.878 ]

Since ( f'\left(\frac{\pi}{6}\right) ) is negative, ( f(x) ) is decreasing at ( x = \frac{\pi}{6} ).

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