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

1 Answer
Steve M
Mar 18, 2018

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 }

Related questions
See all questions in Chain Rule
Impact of this question
1795 views around the world
You can reuse this answer
Creative Commons License