Solving Partial Derivatives: Is This Right?

Does that answer your question?In summary, the conversation discusses a function with two cases and the use of the product and chain rules in finding its partial derivatives. The person asking for help is unsure if they are using the rules correctly and seeks clarification on their approach. The expert confirms that the person's approach is correct and offers additional advice on using the chain rule. They also clarify that the definition of the derivative does not need to be used in the second case.
  • #1
Telemachus
835
30

Homework Statement


Hi there. Well, I got the next function, and I'm trying to work with it. I wanted to know if this is right, I think it isn't, so I wanted your opinion on this which is always helpful.

[tex]f(x,y)=\begin{Bmatrix} (x+y)^2\sin(\displaystyle\frac{\pi}{x+y}) & \mbox{ si }& y\neq-x\\0 & \mbox{si}& y=-x\end{matrix}[/tex]


If [tex]y\neq-x[/tex]:
[tex]\displaystyle\frac{\partial f}{\partial x}=2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{(x+y)})[/tex]

[tex]\displaystyle\frac{\partial f}{\partial y}=2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{(x+y)})[/tex]

Second case:

If [tex]y=-x[/tex]
[tex]f(x,y)=0[/tex]

[tex]\displaystyle\frac{\partial f}{\partial y}=\displaystyle\frac{\partial f}{\partial x}=0[/tex]

Is this right? I would like to know why it isn't, I think its not. Should I think when I consider the second case as I would go in all directions or something like that? cause in other cases where I got defined different functions for a particular point I've used the definition, but in this case I got trajectories. What should I do?

Bye and thanks.
 
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  • #2
Telemachus said:

Homework Statement


Hi there. Well, I got the next function, and I'm trying to work with it. I wanted to know if this is right, I think it isn't, so I wanted your opinion on this which is always helpful.

[tex]f(x,y)=\begin{Bmatrix} (x+y)^2\sin(\displaystyle\frac{\pi}{x+y}) & \mbox{ si }& y\neq-x\\0 & \mbox{si}& y=-x\end{matrix}[/tex]


If [tex]y\neq-x[/tex]:
[tex]\displaystyle\frac{\partial f}{\partial x}=2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{(x+y)})[/tex]

[tex]\displaystyle\frac{\partial f}{\partial y}=2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{(x+y)})[/tex]

Second case:

If [tex]y=-x[/tex]
[tex]f(x,y)=0[/tex]

[tex]\displaystyle\frac{\partial f}{\partial y}=\displaystyle\frac{\partial f}{\partial x}=0[/tex]

Is this right? I would like to know why it isn't, I think its not. Should I think when I consider the second case as I would go in all directions or something like that? cause in other cases where I got defined different functions for a particular point I've used the definition, but in this case I got trajectories. What should I do?
You're not using the product rule correctly in either partial. With ordinary derivative, if h(x) = f(x)*g(x), h'(x) = f'(x)*g(x) + f(x)*g'(x). It's similar for your two partials.

Also, when you differentiate sin(pi/(x + y)), you're not using the chain rule correctly. You will get cos(pi/(x + y)), but now you need the derivative of the argument to the cosine function, so you need the derivative of pi/(x + y).
 
  • #3
Why not?

[tex](x+y)^2\sin(\displaystyle\frac{\pi}{x+y})[/tex]

[tex]\displaystyle\frac{\partial f}{\partial x}=2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{(x+y)})[/tex]

The derivative on the left is obvious. For the right part, I've need to find derivative of [tex]\sin(\displaystyle\frac{\pi}{x+y})[/tex], which is the derivative of the sine multiplied by the derivative on the inside of the sine (Im working with respect to x):
[tex]\cos(\displaystyle\frac{\pi}{x+y})(\displaystyle\frac{-\pi}{(x+y)^2})[/tex]
Then, when I make the product with [tex](x+y)^2[/tex] I got [tex]-\pi\cos(\displaystyle\frac{\pi}{x+y})[/tex]

I don't see what's wrong with it.
 
  • #4
You're right. I didn't work the problems through, but it seemed that you were missing a factor on the second terms of your partials. Your work didn't show that intermediate step (before canceling), so I assumed you had not used the chain rule correctly.
 
  • #5
Thanks. Why I have to use the definition of derivative in the second case?
 
  • #6
No, I don't see why you would need to use the definition of the derivative. Since f(x, y) = 0 along the line y = -x, both partials are zero as well.
 

FAQ: Solving Partial Derivatives: Is This Right?

1. How do I solve partial derivatives?

To solve partial derivatives, you need to take the derivative of a function with respect to one of its variables while holding the other variables constant. This means you will have multiple derivatives for each variable in the function.

2. What is the purpose of solving partial derivatives?

Solving partial derivatives allows us to understand how a function changes with respect to each of its variables. It is useful in many fields of science and engineering, such as physics, economics, and engineering.

3. How do I know if my solution for a partial derivative is correct?

To check if your solution is correct, you can use the rules of partial derivatives to simplify your answer and see if it matches the function you started with. You can also use online calculators or ask a colleague to verify your solution.

4. Are there any strategies for solving partial derivatives?

Yes, there are several strategies that can help make solving partial derivatives easier. These include using the product rule, quotient rule, chain rule, and power rule. It is also helpful to practice and become familiar with common derivatives.

5. Can partial derivatives be solved for any type of function?

Yes, partial derivatives can be solved for any type of function as long as it is differentiable. This means that the function must have a defined derivative for each of its variables. However, some functions may be more challenging to solve and may require more advanced techniques.

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