How do you differentiete g(x)=#e^(2x)(2x+1)# ?

Question

560 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-15T10:41:22

#g'(x)=4e^(2x)(x+1)#

Here,

#g(x)=e^(2x)(2x+1)#

Diff.w.r.t #x# ,#" using "color(blue)"Product rule :"#

#g'(x)=e^(2x)d/(dx)(2x+1)+(2x+1)d/(dx)(e^(2x))#

#=>g'(x)=e^(2x)*2+(2x+1)e^(2x)*2#

#=>g'(x)=2e^(2x)[1+2x+1]#

#=>g'(x)=2e^(2x)(2x+2)#

#=>g'(x)=4e^(2x)(x+1)#