Problemddxtan(x)\frac{d}{dx}\tan(x)dxdtan(x)分步解答改写为 tanx=sinxcosx\tan x = \frac{\sin x}{\cos x}tanx=cosxsinx。应用商的求导法则:(fg)′=f′g−fg′g2\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}(gf)′=g2f′g−fg′。当 f=sinxf = \sin xf=sinx(f′=cosxf' = \cos xf′=cosx)且 g=cosxg = \cos xg=cosx(g′=−sinxg' = -\sin xg′=−sinx)时:cosx⋅cosx−sinx⋅(−sinx)cos2x=cos2x+sin2xcos2x\frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}cos2xcosx⋅cosx−sinx⋅(−sinx)=cos2xcos2x+sin2x。使用毕达哥拉斯恒等式 sin2+cos2=1\sin^2 + \cos^2 = 1sin2+cos2=1:=1cos2x=sec2x= \frac{1}{\cos^2 x} = \sec^2 x=cos2x1=sec2x。答案sec2(x)\sec^2(x)sec2(x)想解其他题?打开 derivative 求解器 →相关例题/solve/calculus/derivative-of-sin-2x/solve/calculus/derivative-of-cos-x