数学解题，大学高数求导

Question

y=lnsinx
y=sin(x+sinx)
y=1/(√1+x²)
y=sin²(2x–1)
y=lnsin(1+x²)
y=lntan(x/2)

Answer 1

[lnsinx]' = 1/sin(x)*(sinx)' = cosx/sinx = ctan(x)
[sin(x+sinx)]' = cos(x + sinx)*[x + sinx]' = [1 + cosx]cos(x+sinx)
[1/(1+x^2)^(1/2)]' = [(1+x^2)^(-1/2)]' = (-1/2)[1+x^2]^(-3/2)*[1+x^2]' = -x/(1+x^2)^(-3/2)

[sin²(2x–1)]' = 2sin(2x-1)*[sin(2x-1)]' = 2sin(2x-1)cos(2x-1)*(2x-1)' = 4sin(2x-1)cos(2x-1) = 2sin(4x-2)

[lnsin(1+ x^2)]' = 1/sin(1+x^2)*[sin(1+x^2)]' = cos(1+x^2)/sin(1+x^2)*[1+x^2]' = 2xcos(1+x^2)/sin(1+x^2) = 2xctan(1+x^2)

[lntan(x/2)]' = 1/tan(x/2)*[tan(x/2)]' = [sec(x/2)]^2/tan(x/2)*[x/2]' = [sec(x/2)]^2/[2tan(x/2)] = 1/[2sin(x/2)cos(x/2)] = 1/sin(x)本回答被提问者采纳