Question

logarithmic differentiation
how do i differentiate something like:
y = 2x^e^x

Answer 1

\[\large y= 2x^{e^x}\]

Answer 2

take nature log for both sides and use implicit differrentication

Answer 3

\[\ln y = e^x \ln 2x\]

Answer 4

yep

Answer 5

\[
\large y= 2x^{e^x}\\
\ln y = \ln 2 + e^x \ln x
\]

Answer 6

then with respect to x, uh, and using the product rule..
\[\frac{ 1 }{ y }y' = \frac{ 1 }{ x } e^x + e^x \ln 2x\]

Answer 7

Yes, now find y'

Answer 8

\[y' = y(\frac{ e^x }{ x } + e^x \ln x)\]

Answer 9

that should be a ln 2x*

Answer 10

Replace y by its original formula

Answer 11

\[\huge y' = 2x^{e^x}(\frac{ e^x }{ x } + e^x \ln 2x)\]