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

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Answer 1 · 2018-02-20T16:15:16

#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

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)#

#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)#