反函数求导

求函数y=tan⁡axy=\tan axy=tanax的反函数x=x(y)x=x(y)x=x(y)的导数。

解：∵\qquad \because∵函数y=tan⁡axy=\tan axy=tanax严格单调递增且可导

\qquad且y ′ = acos ⁡ 2axy'=\dfrac{a}{\cos^2 ax}y′=cos2axa​

∴x=x(y)\qquad \therefore x=x(y)∴x=x(y)可导

x ′ (y)= 1y′(x)=cos2ax a\qquad x'(y)=\dfrac{1}{y'(x)}=\dfrac{cos^2ax}{a}x′(y)=y′(x)1​=acos2ax​

∵ cos ⁡2 ax(1+ tan ⁡2 ax)= cos ⁡2 ax+ sin ⁡2 ax=1\qquad \because \cos^2ax(1+\tan^2 ax)=\cos^2ax+\sin^2 ax=1∵cos2ax(1+tan2ax)=cos2ax+sin2ax=1

∴ cos ⁡2 ax= 11+tan ⁡ 2ax\qquad \therefore \cos^2ax=\dfrac{1}{1+\tan^2 ax}∴cos2ax=1+tan2ax1​

∴x ′ (y)=cos ⁡ 2ax a= 1a+atan ⁡ 2ax= 1a+ay2\qquad \therefore x'(y)=\dfrac{\cos^2ax}{a}=\dfrac{1}{a+a\tan^2 ax}=\dfrac{1}{a+ay^2}∴x′(y)=acos2ax​=a+atan2ax1​=a+ay21​

求函数y=x+e x y=x+e^xy=x+ex的反函数x=x(y)x=x(y)x=x(y)的导数。

解：∵\qquad \because∵函数y=x+e x y=x+e^xy=x+ex严格单调递增且可导

\qquad且y ′ =1+e x y'=1+e^xy′=1+ex

∴x ′ (y)= 1y′(x)= 11+ex\qquad \therefore x'(y)=\dfrac{1}{y'(x)}=\dfrac{1}{1+e^x}∴x′(y)=y′(x)1​=1+ex1​