y = e arctan 2 x y=e^{\arctan \sqrt2 x} y=earctan2x，求 d y dy dy.
解： d y = e arctan 2 x ⋅ 1 1 + 2 x 2 ⋅ 2 d x = 2 e arctan 2 x 1 + 2 x 2 d x dy=e^{\arctan \sqrt2 x}\cdot \dfrac{1}{1+2x^2}\cdot \sqrt2dx=\dfrac{\sqrt2 e^{\arctan \sqrt2 x}}{1+2x^2}dx dy=earctan2x⋅1+2x21⋅2dx=1+2x22earctan2xdx
设 f ( x ) = x ln x f(x)=x\ln x f(x)=xlnx，求 d f ( 2 x ) df(2x) df(2x).
解： d f ( 2 x ) = f ′ ( 2 x ) ⋅ 2 d x = ( ln 2 x + 1 ) ⋅ 2 d x = ( 2 ln 2 x + 2 ) d x df(2x)=f'(2x)\cdot 2dx=(\ln 2x+1)\cdot 2dx=(2\ln 2x+2)dx df(2x)=f′(2x)⋅2dx=(ln2x+1)⋅2dx=(2ln2x+2)dx