数学解题,大学高数求导y=lnsinxy=sin(x+sinx)y=1/(√1+x²)y=sin²(2x–1)y=lnsin(1+x²)y=lntan(x/2)
2019-05-23
数学解题,大学高数求导
y=lnsinx
y=sin(x+sinx)
y=1/(√1+x²)
y=sin²(2x–1)
y=lnsin(1+x²)
y=lntan(x/2)
优质解答
[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)^...
[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)^...