Question

求函数f(x)=xe^x+Inx/sinx的导数

RT速求~~~
并求x=π/2处的切线方程

Answer 1

f'(x)=e^x+xe^x+(sinx/x -lnxcosx)/(sinx)^2
当x=π/2时，
f(π/2)=π/2e^(π/2)+ln(π/2) →切点的纵坐标
f'(π/2)=e^(π/2)(π/2+1)+2/π →切线的斜率
切线方程为y-f(π/2)=f'(π/2)(x-π/2) →点斜式
你自己带入计算就可得出切线方程。

Answer 2

syms x
f=x*exp(x)+log(x)/sin(x)

f =

log(x)/sin(x) + x*exp(x)

diff(f,x)

ans =

exp(x) + 1/(x*sin(x)) + x*exp(x) - (cos(x)*log(x))/sin(x)^2

Answer 3

f(x)'=e^x+xe^x+(sinx/x-lnxcosx)/(sinx)^2
=(1+x)e^x+(sinx-xlnxcosx)/(x(sinx)^2)

Answer 4

f(x)=xe^x+Inx/sinx
f'(x)=e^x+xe^x+1/(xsinx)-cosxlnx/(sinx)^2