How do you find the derivative of # 2e^(x+3)#?

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Answer 1 · 2017-01-17T01:55:13

#f(x)=2e^(x+3) => f'(x)=2e^(x+3)#

First remember a derivative times a constant equals a constant times a derivative

#(cf(x))'=cf'(x)# in this case #c=2#

So

#(2e^(x+3))'=2(e^(x+3))'#

Then we use the chain rule #(f(g(x))'=f'(g(x)g'(x)#

#2(e^(x+3))'=2(e^(x+3))'(x+3)'#

Since #(e^x)'=e^x#

#2(e^(x+3))'(x+3)'=2(e^(x+3))(x+3)'#

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

#2(e^(x+3))(x+3)'=2(e^(x+3))((x)'+(3)')#

Since 3 is a constant its derivative is zero

#2(e^(x+3))((x)'+(3)')=2(e^(x+3))((x)'+0)=2(e^(x+3))(x)'#

and since #(f(x)=x => f'(x)=1#)

#2(e^(x+3))(x)'=2(e^(x+3))(1)=underline(2(e^(x+3)))#