Question

Find the derivative of the function y=(2x)^ln(x), please?

Answer 1

Hey :)
Ooo this one is a doozy!

Answer 2

We could do this:\[\Large\rm y=2^{\ln x}\cdot x^{\ln x}\]and then product rule.
Nahhh let's not do that. That seems tedious.

Answer 3

Exactly :/

Answer 4

Let's just take the log of each side,\[\Large\rm \ln y=\ln x\cdot \ln(2x)\]

Answer 5

How bout that step, that one make sense? :o

Answer 6

I took natural log of each side,
then applied my exponent rule on the right side.

Answer 7

Ooh wait, let me work it out first

Answer 8

Can I use product rule on the right side?

Answer 9

Mmm yes, very good.

Answer 10

I got dy/dx = y(2 ln(2x))/x). Is it correct..?

Answer 11

\[\Large\rm \ln y=\ln x\cdot \ln(2x)\]\[\Large\rm \frac{1}{y}y'=(\ln x)'\ln(2x)+\ln x(\ln2x)'\]I feel like something went wrong on the right side. Hmm.

Answer 12

Don't forget your chain rule when differentiating the ln2x.\[\Large\rm \left(\ln2x\right)'=\frac{1}{2x}(2x)'=\frac{1}{2x}2=\frac{1}{x}\]

Answer 13

\[\Large\rm y'=y\left[\frac{1}{x}\ln(2x)+\ln x\frac{1}{x}\right]\]Something like that, yah? :o

Answer 14

Oops wrong answer haha

Answer 15

Trying to work other ways now :/

Answer 16

Other ways to solve it? :o
You could do this approach that we talked about that the start,\[\Large\rm y=2^{\ln x}\cdot x^{\ln x}\]Then,\[\Large\rm y'=\left(2^{\ln x}\right)'x^{\ln x}+2^{\ln x}\left(x^{\ln x}\right)'\]

Answer 17

But then you have to separately figure out \(\Large\rm \left(x^{\ln x}\right)'\) which is a bit of a pain :d
It's like a problem within a problem lol

Answer 18

Lol I knoww

Answer 19

\[\text{Let }g = x^{\ln x}\\~\\\ln g = (\ln x)^2\\~\\ \dfrac{g'}{g} = \dfrac{2\ln x}{x}\\~\\g' = g\left[\dfrac{2\ln x}{x}\right] = x^{\ln x}\dfrac{2\ln x}{x} = \left(x^{\ln x}\right)'\]
Looks like tedious, but not really tedious. But again I have worked with calculus problems a lot...

Answer 20

the problem is the answer can be formatted a few different ways.
with
\[ y = \left(2x\right)^{\ln x} \\ \ln(y) = \ln x \ln(2x) \\
\frac{\dot{y}}{y} = \ln x \frac{1}{2x} \cdot 2 + \ln(2x) \cdot \frac{1}{x} \\
\dot{y}= \left(2x\right)^{\ln x} \left( \frac{\ln x + \ln(2x)}{x}\right)
\]
However, we can write this many ways:
\[ \dot{y}= 2^{\ln x} x^{\ln x -1 } ( \ln x + \ln 2x) \\
\dot{y}= 2^{\ln x} x^{\ln x -1 } \ln(2x^2) \\
\dot{y}= 2^{\ln x} x^{\ln x -1 } (\ln(x^2)+\ln(2) )\\
\dot{y}= 2^{\ln x} x^{\ln x -1 } (2\ln(x)+\ln(2) )
\]
and we can continue tweaking the form...

Answer 21

Thank youuuuu !