# Simple but i cant get it? Log derivative

1. May 19, 2005

### Struggling

hi guys, im new to forums and need help cause im doing engineering 1st year, and im horrible at maths and chemistry (bad match )

equation is simple,

x^2 Log x = 1,

i have no idea wat to do with the 1

i turned it into x^2 log x -1 = 0 and then used y'=uv' + vu' and got:
x log(x-1)
but i dont think its right because when i use newtons method it comes out messed up.
any help would be appreciated.

thanks

2. May 19, 2005

### arildno

The first thing you need to learn, is to state your problem CLEARLY, both for yourself and others.

Is your problem to devise an algorithm (based on Newton's method) for numerical computation of the solution of this equation?

3. May 19, 2005

### Struggling

question:

Use newtons method to locate the root of x^2 log x =1 correct to three decimal places.

f(x) = x^2 log x = 1
f '(x) = ??????????

is that what you mean?

4. May 19, 2005

### arildno

First of all, you need to learn what a "root" is in this context.
Given som function H(x), we say that a number $$x^{*}$$ is a ROOT of H if we have $$H(x^{*})=0$$

Now, by looking at your equation, what is the function you are to find the root to?

5. May 19, 2005

### Zurtex

O.K so:

$$f(x) = x^2 \log x - 1$$

Using the product rule:

$$f'(x) = x^2 \frac{1}{x} + 2 x \log x = x + 2x \log x = x(1 + 2 \log x)$$

Which means your itterative forumulae for the Newton-Raphson should look something like:

$$x_i = x_{i-1} - \frac{f(x_{i-1})}{f'(x_{i-1})}$$

6. May 19, 2005

### cronxeh

Finding a root means finding the value of x (independent variable) when y (dependent variable) is 0.

Whenever you have a problem like that you should always plot it, the answer lies between 1.4 and 1.6 according to my plot

Last edited: Oct 8, 2005
7. May 19, 2005

### Struggling

thank you all very much, i found the mistake in the product rule

8. Jun 11, 2005

### Struggling

sorry to bring this back up again i was just going over my notes and its better than restarting the thread.
using the product rule y' = uv'+vu'

shouldnt the equation be x^2(1/x-1) + (log x-1)(2x)??????

dat means it would be y' = -x + 2x log x - 1 ?????

because derivative of a log is log x = 1/x

im just a little confused what happens with the 1

9. Jun 11, 2005

### Nylex

The derivative of a constant is 0.

10. Jun 11, 2005

### Zurtex

It's quite simple :

$$\frac{d}{dx} \left( x^2 \log x - 1 \right) = \frac{d}{dx} \left( x^2 \log x \right) + \frac{d}{dx}(-1) = \frac{d}{dx} \left( x^2 \log x \right) + 0$$

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