Solving d/dx from dy/dx in Maths

  • Thread starter htdIO
  • Start date
In summary, the conversation is discussing the application of the chain rule and product rule in simplifying the expression (\frac{1}{x})(\frac{dy}{dx}) - (\frac{1}{x^2})y, which can also be written as (\frac{d}{dx})[(\frac{1}{x})y]. The participants are seeking clarification on how to apply these rules and eventually come to the conclusion that the quotient rule is the most efficient method.
  • #1
htdIO
7
0
Hi all,

This is not strictly a DE question, but I came across this while working on one. This isn't the first time I got this and I just can't remember this from my 1st year maths. Some knowledge would be greatly appreciated. In the answer they do the following:

[itex](\frac{1}{x})(\frac{dy}{dx}) - (\frac{1}{x^2})y \Rightarrow

(\frac{d}{dx})[(\frac{1}{x})y][/itex]

Now I want to know how? I just can't simplify it. Silly question, but need the help!

Thanks
 
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  • #2
Welcome to PF, htdIO! :smile:

Are you familiar with the chain rule?

It is: [itex]\frac d {dx} f(y(x)) = \frac {df} {dy} \frac {dy} {dx}[/itex]

Do you know how to apply this?
 
  • #3
Hi and thanks!

I do know it. Just not quite sure how I should be applying it here? I've scribbled quite a bit down here now, trying to combine this with the product rule. Or am I heading in the wrong direction?
 
  • #4
Sorry, you're right. You need to apply the product rule.
Do you know how to apply it to: [itex](\frac{d}{dx})[(\frac{1}{x})y(x)][/itex]?
 
  • #5
Haha, aah thanks. I must be more tired than I thought...
I'm guessing the only way to 'see' this (like they did it), is by recognizing it and a bit of practice?
 
  • #6
Hah, after all the practice I got, I thought you needed the chain rule!
So much for all that practice! :wink:
 
  • #7
Halfway through I actually remembered the quotient rule, which should make it quicker ;) Anyway, thanks again for getting me on the right track!
 
  • #8
Neh, the quotient rule is not quicker in this case.
But good you remembered it! :smile:
 

FAQ: Solving d/dx from dy/dx in Maths

What is a derivative?

A derivative is a mathematical concept that represents the rate of change of one variable with respect to another variable. In terms of functions, it is the slope of the tangent line at a specific point on the graph of the function.

What is the difference between d/dx and dy/dx?

d/dx is the notation used to represent the derivative of a function with respect to the independent variable x. dy/dx, on the other hand, represents the derivative of the dependent variable y with respect to the independent variable x.

Why is it important to solve for d/dx from dy/dx?

Solving for d/dx from dy/dx allows us to find the equation of the tangent line at a specific point on the graph of a function. This is useful in calculating rates of change, finding maximum and minimum values, and understanding the behavior of a function at a particular point.

What are the different methods for solving d/dx from dy/dx?

There are several methods for solving d/dx from dy/dx, including the power rule, product rule, quotient rule, and chain rule. Each method is used for different types of functions and allows us to find the derivative of more complex functions by breaking them down into simpler parts.

How can I check if my solution for d/dx from dy/dx is correct?

One way to check if your solution is correct is by taking the derivative of the original function using the methods mentioned above. If your solution matches the derivative, then it is correct. Another way is to use graphing software to plot the original function and the tangent line at the specific point to visually confirm if the slope of the tangent line matches your solution.

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