y = y ( x ) y=y(x) y=y(x)是由方程 e x − e y + 1 = cos ( x y ) e^x-e^y+1=\cos(xy) ex−ey+1=cos(xy)所确定的函数，求 d y d x \dfrac{dy}{dx} dxdy.
解：
\qquad
两边同时求导得：
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\qquad e^x-e^yy'=-\sin(xy)\times(y+xy')
ex−eyy′=−sin(xy)×(y+xy′)
( x sin ( x y ) − e y ) y ′ = − y sin ( x y ) − e x \qquad (x\sin(xy)-e^y)y'=-y\sin(xy)-e^x (xsin(xy)−ey)y′=−ysin(xy)−ex
y ′ = e x + y sin ( x y ) e y − x sin ( x y ) \qquad y'=\dfrac{e^x+y\sin(xy)}{e^y-x\sin(xy)} y′=ey−xsin(xy)ex+ysin(xy)
\qquad 所以 d y d x = y ′ = e x + y sin ( x y ) e y − x sin ( x y ) \dfrac{dy}{dx}=y'=\dfrac{e^x+y\sin(xy)}{e^y-x\sin(xy)} dxdy=y′=ey−xsin(xy)ex+ysin(xy)
设 y = ( 1 + x 2 ) ( 1 + 2 x ) ( 1 + x ) 3 ( 1 − 2 x ) 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：
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\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)
= 1 2 ln ( 1 + x 2 ) + 1 2 ln ∣ 1 + 2 x ∣ − 3 2 ln ∣ 1 + x ∣ − 1 2 ln ∣ 1 − 2 x ∣ \qquad\qquad=\dfrac 12\ln(1+x^2)+\dfrac 12\ln|1+2x|-\dfrac 32\ln|1+x|-\dfrac 12\ln|1-2x| =21ln(1+x2)+21ln∣1+2x∣−23ln∣1+x∣−21ln∣1−2x∣
\qquad 两边同时求导得：
y ′ y = x 1 + x 2 + 1 1 + 2 x − 3 2 ( 1 + x ) + 1 1 − 2 x \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 + x 2 ) ( 1 + 2 x ) ( 1 + x ) 3 ( 1 − 2 x ) [ x 1 + x 2 + 1 1 + 2 x − 3 2 ( 1 + x ) + 1 1 − 2 x ] \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]