若 y = f ( u ) y=f(u) y=f(u)， u = g ( x ) u=g(x) u=g(x)，则 d y = f ′ ( u ) d u = f ′ ( u ) ⋅ g ′ ( x ) d x dy=f'(u)du=f'(u)\cdot g'(x)dx dy=f′(u)du=f′(u)⋅g′(x)dx
形式不变性指若 y = f ( u ) y=f(u) y=f(u)，则无论 u u u是自变量还是中间变量，对 y y y微分保持形式不变， d y = f ′ ( u ) d u dy=f'(u)du dy=f′(u)du
y = cos ln ( 2 x + 1 ) y=\cos\ln(2x+1) y=cosln(2x+1)，求 d y dy dy.
解： d y = − sin ln ( 2 x + 1 ) ⋅ 1 2 x + 1 ⋅ 2 d x = − 2 sin ln ( 2 x + 1 ) 2 x + 1 d x dy=-\sin\ln(2x+1)\cdot\dfrac{1}{2x+1}\cdot 2dx=-\dfrac{2\sin\ln(2x+1)}{2x+1}dx dy=−sinln(2x+1)⋅2x+11⋅2dx=−2x+12sinln(2x+1)dx
设 f ( x ) f(x) f(x)可导， y = f ( sin x ) + e f ( x ) y=f(\sin x)+e^{f(x)} y=f(sinx)+ef(x)，求 d y dy dy.
解： d y = [ f ′ ( sin x ) ⋅ cos x + e f ( x ) ⋅ f ′ ( x ) ] d x dy=[f'(\sin x)\cdot \cos x+e^{f(x)}\cdot f'(x)]dx dy=[f′(sinx)⋅cosx+ef(x)⋅f′(x)]dx