How do you differentiate #f(x)=(e^x-3lnx)(tanx+2x)# using the product rule?

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Answer 1 · 2016-01-05T00:58:09

#f'(x)=(e^x-3/x)(tanx+2x)+(e^x-3lnx)(sec^2x+2)#

The product rule states that for a function #f(x)=g(x)h(x)#,

#f'(x)=g'(x)h(x)+h'(x)g(x)#

Thus

#f'(x)=(tanx+2x)d/dx(e^x-3lnx)+(e^x-3lnx)d/dx(tanx+2x)#

Find the derivative of each part separately.

#d/dx(e^x-3lnx)=e^x-3/x#

#d/dx(tanx+2x)=sec^2x+2#

Hence

#f'(x)=(e^x-3/x)(tanx+2x)+(e^x-3lnx)(sec^2x+2)#