How do you differentiate #f(x)=(lnx-x)(cotx-2e^x)# using the product rule?

1 Answer
Oct 31, 2016

Answer:

# f'(x) = (cotx-2e^x)(1/x-1) - (lnx-x)(csc^2x+2e^x) #

Explanation:

I will assume you are familiar with these standard differenmtiation result:
# d/dx(lnx)=1/x#; #d/dx(e^x)=e^x#; #d/dx(cotx)=-csc^2x#

The product rule is # d/dx(uv)=u(dv)/dx+v(du)/dx #

so if # f(x) = (lnx-x)(cotx-2e^x) #; then

# f'(x) = (lnx-x)(d/dx(cotx-2e^x)) + (cotx-2e^x)(d/dx(lnx-x)) #

# :. f'(x) = (lnx-x)(-csc^2x-2e^x) + (cotx-2e^x)(1/x-1) #

# :. f'(x) = (lnx-x)(-1)(csc^2x+2e^x) + (cotx-2e^x)(1/x-1) #

# :. f'(x) = (cotx-2e^x)(1/x-1) - (lnx-x)(csc^2x+2e^x) #