Simple but i cant get it? Log derivative

In summary, the conversation discusses a problem with an equation involving logarithms and Newton's method. The participants offer suggestions and clarification on how to solve the problem and find the root of the equation. They also discuss the correct use of the product rule in finding the derivative of the equation.
  • #1
Struggling
52
0
hi guys, I am new to forums and need help cause I am doing engineering 1st year, and I am horrible at maths and chemistry (bad match :confused: )

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 don't think its right because when i use Newtons method it comes out messed up.
any help would be appreciated.

thanks
 
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  • #2
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
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
Struggling said:
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?
First of all, you need to learn what a "root" is in this context.
Given som function H(x), we say that a number [tex]x^{*}[/tex] is a ROOT of H if we have [tex]H(x^{*})=0[/tex]

Now, by looking at your equation, what is the function you are to find the root to?
 
  • #5
O.K so:

[tex]f(x) = x^2 \log x - 1[/tex]

Using the product rule:

[tex]f'(x) = x^2 \frac{1}{x} + 2 x \log x = x + 2x \log x = x(1 + 2 \log x)[/tex]

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

[tex]x_i = x_{i-1} - \frac{f(x_{i-1})}{f'(x_{i-1})}[/tex]
 
  • #6
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:
  • #7
thank you all very much, i found the mistake in the product rule :biggrin:
 
  • #8
Zurtex said:
O.K so:

[tex]f(x) = x^2 \log x - 1[/tex]

Using the product rule:

[tex]f'(x) = x^2 \frac{1}{x} + 2 x \log x = x + 2x \log x = x(1 + 2 \log x)[/tex]

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

[tex]x_i = x_{i-1} - \frac{f(x_{i-1})}{f'(x_{i-1})}[/tex]


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
Struggling said:
im just a little confused what happens with the 1

The derivative of a constant is 0.
 
  • #10
Struggling said:
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
It's quite simple :smile: :

[tex]\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[/tex]
 

Related to Simple but i cant get it? Log derivative

1. What is a simple log derivative?

A simple log derivative is a mathematical concept used to find the derivative of a logarithmic function. It involves using the properties of logarithms and the chain rule to determine the rate of change of a logarithmic function.

2. How do I find the log derivative of a function?

To find the log derivative of a function, you first need to rewrite the function as a logarithm. Then, you can use the properties of logarithms and the chain rule to find the derivative. This may involve taking the natural logarithm (ln) or the base 10 logarithm (log) of the function.

3. Why is the log derivative important?

The log derivative is important because it allows us to find the rate of change of a logarithmic function, which can be useful in various real-world applications. It also helps us to simplify complex functions and make them easier to work with.

4. What is the difference between a log derivative and a regular derivative?

The difference between a log derivative and a regular derivative is that a log derivative is specifically used for logarithmic functions, while a regular derivative can be used for any type of function. The process of finding a log derivative involves using the properties of logarithms, while finding a regular derivative involves using the basic rules of differentiation.

5. Can you give an example of finding a log derivative?

Sure, let's say we have the function f(x) = ln(x^2). To find the log derivative, we first rewrite the function as f(x) = 2ln(x). Then, we use the power rule and the chain rule to find the derivative, which would be f'(x) = 2/x. This means that the log derivative of ln(x^2) is 2/x.

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