f'（cosx+2）=sin2x+tan2x求f(x)

f'（cosx+2）=sin2x+tan2x求f(x)
f'（cosx+2）=sin2x+tan2x求f(x)

f'（cosx+2）=sin2x+tan2x求f(x)
令cosx+2=t
则cosx=t-2
∴san²x=1-（t-2）²,tan²x=sin²x/cos²x=1/(t-2)² -1
所以代入f′（t）=1/(t-2)²-（t-2）²
由1/x的导数为-1/x² 可推出 1/(t-2)²对应的原函数为-1/（t-2)
(t-2)²对应原函数为1/3·(t-2)³
∴f（t）=-1/(t-2) -1/3·（t-2）³
最后将t 换成x即可

f'(cosx+2)=[cos(2x)+1]/2+1/cos^2x-1=cos(2x)/2+(tanx)'-1/2=[sin(2x)/4]'+(tanx)'-[x/2]'
=[sin(2x)/4+tanx-x/2]'=[sinxcosx/2+sinx/cosx-x/2]'=[sinx(cos^2x+2)/(2cosx)-x/2]'
3>=cosx+2=t>=1,x=arccos(t...

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f'(cosx+2)=[cos(2x)+1]/2+1/cos^2x-1=cos(2x)/2+(tanx)'-1/2=[sin(2x)/4]'+(tanx)'-[x/2]'
=[sin(2x)/4+tanx-x/2]'=[sinxcosx/2+sinx/cosx-x/2]'=[sinx(cos^2x+2)/(2cosx)-x/2]'
3>=cosx+2=t>=1,x=arccos(t-2),cosx=t-2,cos^2x=(t-2)^2,sinx=|√[1-(t-2)^2]|
f'(t)=[sinx(cos^2x+2)/(2cosx)-x/2]'={|√[1-(t-2)^2]|[(t-2)^2+2]/[2(t-2)]-arccos(t-2)/2}'
f(t)=|√[1-(t-2)^2]|[(t-2)^2+2]/[2(t-2)]-arccos(t-2)/2+C

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