넓이 함수의 미분

두 함수 $y = \sin x$, $y = a \tan x$의 교점의 $x$좌표를 $\displaystyle \alpha \left( 0 < \alpha < \frac{\pi}{2} \right)$라 하면 $\sin \alpha = a \tan \alpha$, $\displaystyle \sin \alpha = a \cdot \frac{\sin \alpha}{\cos \alpha}$ $\sin \alpha \neq =$이므로 $\cos \alpha = a$ $f \left( a \right)$$\displaystyle = \int _{0} ^{\alpha} \left( \sin x - a \tan x \right) dx$ $\displaystyle = \int _{0} ^{\alpha} \sin x dx - a \int _{0} ^{\alpha} \tan x dx$ $\cdots \cdots$ ㉠ ㉠을 $a$에 대하여 미분하면 $f' \left( a \right)$$\displaystyle = \sin \alpha \frac{d \alpha}{da} - \int _{0} ^{\alpha} \tan x dx - a \tan \alpha \frac{d \alpha}{da}$ $\displaystyle = \left( \sin \alpha - \cos \alpha \tan \alpha \right) \frac{d \alpha}{da} - \frac{\int _{0} ^{\alpha {\sin x}}}{\cos x} dx$ $\displaystyle = \left[ \begin{array}{l} \ln \cos x \begin{matrix} \\ \end{matrix} \end{array} \right] _{0} ^{\alpha}$ $= \ln \cos \alpha$ $= \ln a$ $\therefore$$\displaystyle f' \left( \frac{1}{e ^{2}} \right) = \ln \frac{1}{e ^{2}} = \mathit{-} 2$