隐函数求导

两边同时对x求导

y=y(x)y=y(x)y=y(x)是由方程ex−ey+1=cos⁡(xy)e^x-e^y+1=\cos(xy)ex−ey+1=cos(xy)所确定的函数，求dydx\dfrac{dy}{dx}dxdy​.

解：
\qquad两边同时求导得：
ex−eyy′=−sin⁡(xy)×(y+xy′)\qquad e^x-e^yy'=-\sin(xy)\times(y+xy')ex−eyy′=−sin(xy)×(y+xy′)

(xsin⁡(xy)−ey)y′=−ysin⁡(xy)−ex\qquad (x\sin(xy)-e^y)y'=-y\sin(xy)-e^x(xsin(xy)−ey)y′=−ysin(xy)−ex

y′=ex+ysin⁡(xy)ey−xsin⁡(xy)\qquad y'=\dfrac{e^x+y\sin(xy)}{e^y-x\sin(xy)}y′=ey−xsin(xy)ex+ysin(xy)​

\qquad所以dydx=y′=ex+ysin⁡(xy)ey−xsin⁡(xy)\dfrac{dy}{dx}=y'=\dfrac{e^x+y\sin(xy)}{e^y-x\sin(xy)}dxdy​=y′=ey−xsin(xy)ex+ysin(xy)​

两边取ln再求导

设y=(1+x2)(1+2x)(1+x)3(1−2x)y=\sqrt{\dfrac{(1+x^2)(1+2x)}{(1+x)^3(1-2x)}}y=(1+x)3(1−2x)(1+x2)(1+2x)​​，求y′y'y′

解：
ln⁡y=ln⁡(1+x2)(1+2x)(1+x)3(1−2x)\qquad\ln y=\ln \sqrt{\dfrac{(1+x^2)(1+2x)}{(1+x)^3(1-2x)}}lny=ln(1+x)3(1−2x)(1+x2)(1+2x)​​

=12ln⁡(1+x2)+12ln⁡∣1+2x∣−32ln⁡∣1+x∣−12ln⁡∣1−2x∣\qquad\qquad=\dfrac 12\ln(1+x^2)+\dfrac 12\ln|1+2x|-\dfrac 32\ln|1+x|-\dfrac 12\ln|1-2x|=21​ln(1+x2)+21​ln∣1+2x∣−23​ln∣1+x∣−21​ln∣1−2x∣

\qquad两边同时求导得：

y′y=x1+x2+11+2x−32(1+x)+11−2x\qquad \dfrac{y'}{y}=\dfrac{x}{1+x^2}+\dfrac{1}{1+2x}-\dfrac{3}{2(1+x)}+\dfrac{1}{1-2x}yy′​=1+x2x​+1+2x1​−2(1+x)3​+1−2x1​

y′=(1+x2)(1+2x)(1+x)3(1−2x)[x1+x2+11+2x−32(1+x)+11−2x]\qquad y'=\sqrt{\dfrac{(1+x^2)(1+2x)}{(1+x)^3(1-2x)}}[\dfrac{x}{1+x^2}+\dfrac{1}{1+2x}-\dfrac{3}{2(1+x)}+\dfrac{1}{1-2x}]y′=(1+x)3(1−2x)(1+x2)(1+2x)​​[1+x2x​+1+2x1​−2(1+x)3​+1−2x1​]