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

1 Answer
Feb 20, 2018

d/dx((lnx+x)(tanx+e^x))=(lnx+x)(e^x+sec^2x)+(tanx+e^x)((tanx+e^x), which can be simplified as necessary

Explanation:

Given:
f(x)=(lnx+x)(tanx+e^x)
Let
y=f(x)

if u=lnx+x, then
(du)/dx=tanx+e^x

if v=tanx+e^x, then
(dv)/dx=sec^2x+e^x
(dv)/dx=e^x+sec^2x

f(x)=uv
y=uv
(dy)/dx=d/dx(y)

d/dx(y)=d/dx(uv)

Thus,
(dy)/dx=d/dx(uv)

By product rule

d/dx(uv)=u(dv)/dx+v(du)/dx
Substituting
d/dx((lnx+x)(tanx+e^x))=(lnx+x)(e^x+sec^2x)+(tanx+e^x)((tanx+e^x)