How do you differentiate #y = (x)^(e^(2x))#?

1 Answer
Mar 18, 2018

Answer:

# dy/dx = e^(2x) \ x^(e^(2x)) \ { 1/x + ln x^2 }#

Explanation:

We have:

# y = x^(e^(2x)) #

We can take natural logarithms and use implicit logarithmic differentiation:

# ln y = ln {x^(e^(2x))} #
# \ \ \ \ \ = e^(2x) \ ln x #

Then we can use the product rule:

# 1/y \ dy/dx = e^(2x) \ 1/x + 2e^(2x) \ ln x #

# :. 1/y \ dy/dx = e^(2x) \ { 1/x + 2ln x }#

# :. dy/dx = y \ e^(2x) \ { 1/x + ln x^2 }#

# :. dy/dx = x^(e^(2x)) \ e^(2x) \ { 1/x + ln x^2 }#

# :. dy/dx = e^(2x) \ x^(e^(2x)) \ { 1/x + ln x^2 }#