How do you differentiate #e^((ln2x)^2+x) # using the chain rule?

1 Answer
Nov 2, 2016

Answer:

# d/dx{e^((ln2x)^2+x)} = ({2ln2x)/x+1)e^((ln2x)^2+x) #

Explanation:

It requires several application of the chain rule , which I will do implicity:

# d/dx{e^((ln2x)^2+x)} = d/dx{{ln2x)^2+x}e^((ln2x)^2+x) #
# d/dx{e^((ln2x)^2+x)} = {d/dx{ln2x)^2+1}e^((ln2x)^2+x) #
# d/dx{e^((ln2x)^2+x)} = {2(ln2x)^1d/dx(ln2x)+1}e^((ln2x)^2+x) #
# d/dx{e^((ln2x)^2+x)} = {2(ln2x)(1/x)+1}e^((ln2x)^2+x) #
# d/dx{e^((ln2x)^2+x)} = ({2ln2x)/x+1)e^((ln2x)^2+x) #