Question

differentiate and simplify
y=x^x-1

Answer 1

take log of both sides?
lny=lnx^x-1

Answer 2

we have to use log. diff. to find

Answer 3

x^x*log(x)+x^x ??

Answer 4

\[\Large \ln y=\ln(x^x-1)\]Hmm that 1 is causing a problem.
We should probably move it to the other side before applying the log.\[\Large \ln(y+1)=\ln(x^x)\]

Answer 5

the answer is x^(x-1)[(1/x)-1-lnx]. just can't get there. the problem is y=x ^(x-1)

Answer 6

Oh I see :)

Answer 7

\[\Large \ln y=\ln(x^{x-1})\]

Applying our rule of logs:\[\Large \color{teal}{\log(a^b)\quad=\quad b\cdot \log(a)}\]

Answer 8

gives us:\[\Large \ln y \quad=\quad (x-1)\ln x\]right?

Answer 9

yes

Answer 10

From this point we can take our derivative, yay!

Answer 11

\[\Large \color{royalblue}{\left(\ln y\right)'}\quad=\quad \color{royalblue}{(x-1)'}\ln x+(x-1)\color{royalblue}{\left(\ln x\right)'}\]

Answer 12

So we need to apply the product rule on the right side, yes?
Do you understand how to take the derivative of the left side?

Answer 13

no

Answer 14

Remember the derivative of ln x?

Answer 15

dy/dx

Answer 16

no no no silly :)

\[\Large (\ln x)' \quad=\quad\frac{1}{x}\]Remember? :D

Answer 17

oh, yeah.

Answer 18

The same thing happens with (ln y)' except we have to remember one important thing.
Since we're taking the derivative with `respect to x` of a `non-x variable`, we need to toss a y' onto it. (which is the same as dy/dx).

Answer 19

\[\Large (\ln y)' \quad=\quad \frac{1}{y}y'\]

Answer 20

i still do not see the answer in there. can you put the steps together?
lny=lnx^(x-1) then...(with the derivatives put in). please

Answer 21

Ya we still have a ways to go before we get to the answer :)
Ok so taking the derivative of both sides gives us this setup:\[\Large \color{royalblue}{\left(\ln y\right)'}\quad=\quad \color{royalblue}{(x-1)'}\ln x+(x-1)\color{royalblue}{\left(\ln x\right)'}\]Calculating the derivative of the left side gave us:\[\Large \color{orangered}{\frac{1}{y}\;y'}\quad=\quad \color{royalblue}{(x-1)'}\ln x+(x-1)\color{royalblue}{\left(\ln x\right)'}\]

Answer 22

Calculate the derivative of the first blue part on the right, what do you get? :)
Derivative of (x-1) = ?

Answer 23

(1/y)(dy/dx)=...

Answer 24

derivative of (x-1) = ? +_+

Answer 25

comeon gal! This is a nice easy one! :) lol

Answer 26

1

Answer 27

yay! good good.\[\Large \color{orangered}{\frac{1}{y}\;y'}\quad=\quad \color{orangered}{(1)}\ln x+(x-1)\color{royalblue}{\left(\ln x\right)'}\]

Answer 28

We discussed the other derivative a lil bit back.
So here it is again, just in case you forgot:\[\Large \color{orangered}{\frac{1}{y}\;y'}\quad=\quad \color{orangered}{(1)}\ln x+(x-1)\color{orangered}{\left(\frac{1}{x}\right)}\]

Answer 29

yep

Answer 30

We're trying to find y'.
So let's multiply both sides by y.
The y and 1/y cancel on the left side, giving us:\[\Large y'\quad=\quad y\left(\ln x+\frac{x-1}{x}\right)\]

Answer 31

We've got just a little bit more work ahead of us.

Answer 32

We want the `right side` to be written in terms of x.
So we need to change the y to x's using the equation we started with.\[\Large \color{teal}{y=x^{x-1}}\]So our derivative function:\[\Large y'\quad=\quad \color{teal}{y}\left(\ln x+\frac{x-1}{x}\right)\]Becomes:\[\Large y'\quad=\quad \color{teal}{x^{x-1}}\left(\ln x+\frac{x-1}{x}\right)\]

Answer 33

okay, so why is (-) in the answer: x^(x-1)[(1/x)-1-lnx]?

Answer 34

Did I miss a negative somewhere? Hmm lemme check.

Answer 35

Hmm the answer shouldn't look like that.
It should be:\[\Large y'\quad=\quad x^{x-1}\left(-\frac{1}{x}+1+\ln x\right)\]

Answer 36

With all the signs reversed inside the brackets ^ like that.
Unless your final answer is showing a (-) outside of the brackets as well.

Answer 37

okay. i will ask teacher tomorrow. thanks! :)

Answer 38

cool c: