Question #bbdc7

3 Answers
Azimet
Feb 1, 2017

#= 6e^(2x)#

Explanation:

We will require the use of the chain rule: If #u# is a function of #x# and #y# is a function of #u#, then:

#(dy)/(dx) = (dy)/(du) (du)/(dx)#

If we let #u = 2x#, then:

#d/dx(3e^(2x)) = d/(du)(3e^u) * d/dx(u)#

#=3e^u * 2 = 6e^u = 6e^(2x)#. (since #d/dx(2x) = 2#)

Note:
If #y# is a function of #x#, then #d/dx(cy) = c(dy)/(dx)#, where #c# is a real, nonzero constant.

Jim G.
Feb 1, 2017

#6e^(2x)#

Explanation:

the standard derivative of #e^x" is " e^x#

To differentiate #e^(2x)" use the" color(blue)" chain rule"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(d/dx(e^(f(x)))=e^(f(x)) .f'(x))color(white)(2/2)|)))#

#rArrd/dx(3e^(2x))=3e^(2x).d/dx(2x)=6e^(2x)#

Jim G.
Feb 1, 2017

see explanation.

Explanation:

#color(orange)"Reminder"#

#ln(xy)=lnx+lny#

#rArrln(3e^(2x))≠2xln(3e)#

#"but "ln(3e^(2x))=ln3+lne^(2x)=ln3+2xcancel(lne)#

#"and "d/dx(ln3+2x)=2#

#rArrm'(x)=2m(x)=2.3e^(2x)=6e^(2x)#

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