Question

cos²x导数是多少

Answer 1

f（X）=cosx×cosx,f'（X）=（-sinxcosx）+（-cosxsinx）=-2sinxcosx

Answer 2

引用前回国好的回答：
令x=asect,dx=sect tant dt
原式=∫atant*d(asect)
=a^2∫tant sect tant dt (1)
=a^2∫tant d sec t
=a^2tant sect -a^2∫sect sec^2 t dt
=a^2tant sect -a^2∫ sect(tan^2 t+1) dt
=a^2tant sect -a^2∫sect dt -a^2∫tant sect tant dt (2)
联立(1),(2)
所以
a^2∫tant sect tant dt=a^2/2[tant sect -∫sect dt]
=a^2/2[tant sect- ln|sect+tan t| ]+C
=a^2/2*x√(x²-a²)-a^2/2ln|x+√(x²-a²)|+c

y=cos²x
令t=cosx
y=t²
y'=2t×t'
y'=2cosx×(cosx)'
y'=2cosx×(-sinx)
y'=-2sinxcosx
y'=-sin2x

Answer 3

令x=asect,dx=sect tant dt
原式=∫atant*d(asect)
=a^2∫tant sect tant dt (1)
=a^2∫tant d sec t
=a^2tant sect -a^2∫sect sec^2 t dt
=a^2tant sect -a^2∫ sect(tan^2 t+1) dt
=a^2tant sect -a^2∫sect dt -a^2∫tant sect tant dt (2)
联立(1),(2)
所以
a^2∫tant sect tant dt=a^2/2[tant sect -∫sect dt]
=a^2/2[tant sect- ln|sect+tan t| ]+C
=a^2/2*x√(x²-a²)-a^2/2ln|x+√(x²-a²)|+c

Answer 4

-sin2x