Solving Partial Derivatives Homework: fx(x,y) and fy(x,y)

In summary, the partial derivatives of f(x,y) are fx(x,y) = 1/(x+y^2) and fy(x,y) = y^2/(x+y^2)^2. The limit should be simplified before taking it, and the correct answer is not 1.
  • #1
KillerZ
116
0

Homework Statement



Use the definition of partial deriviatives as limits to find fx(x,y) and fy(x,y).


Homework Equations



f(x,y) = [tex]\frac{x}{x + y^{2}}[/tex]


The Attempt at a Solution



I don't think this is right because I think I should have an answer of 1.

fx(x,y) = lim h-> 0 [f(x+h,y) - f(x,y)]/h

=lim h->0 [(x+h)/(x+h+y^2) - x/(x+y^2)]/h
=lim h->0 [(x+h)/(x+h+y^2) - x/(x+y^2)]*1/h
=lim h->0 (x+h)/(xh+h^2+(y^2)h) - x/(xh+(y^2)h)
=lim h->0 ((x/h)+1)/(x+h+y^2) - (x/h)/(x+y^2)
=1/(x+y^2)
 
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  • #2
I don't think the answer is 1. Look at your last line. You got things like (x/h). If you take lim h->0, that goes to infinity. You need to do enough algebra to cancel the h in the denominator before you can find a sensible limit. Combine (x+h)/(x+h+y^2) - x/(x+y^2) into single fraction and simplify the numerator before you take the limit.
 
  • #3
fx(x, y) is not equal to 1.

I think you have an error in this line:
=lim h->0 ((x/h)+1)/(x+h+y^2) - (x/h)/(x+y^2)

I don't understand how you got to this expression from the one just before it.

After taking the limit, you should end up with y^2/(x + y^2)^2
 

1. What are partial derivatives and why are they important in scientific calculations?

Partial derivatives are used to measure the rate of change of a function with respect to one of its variables while holding all other variables constant. They are important in scientific calculations because they allow us to analyze how a function changes in multiple dimensions, which is often necessary in physics, engineering, and other scientific fields.

2. How do I solve for partial derivatives?

To solve for partial derivatives, you need to use the chain rule and treat the other variables as constants. Take the derivative of the function with respect to one variable at a time, and plug in the constant values for the other variables. This will give you the partial derivative for that variable.

3. What is the difference between fx(x,y) and fy(x,y)?

fx(x,y) is the partial derivative of a function with respect to the variable x, while fy(x,y) is the partial derivative with respect to the variable y. In other words, fx(x,y) measures the rate of change of the function in the x direction, while fy(x,y) measures the rate of change in the y direction.

4. Can I use partial derivatives to find maximum or minimum values of a function?

Yes, you can use partial derivatives to find maximum or minimum values of a function in the same way that you would use regular derivatives. Simply set the partial derivatives equal to 0 and solve for the variables. The resulting values will be the coordinates of the maximum or minimum point.

5. What are some real-world applications of solving partial derivatives?

Partial derivatives have many real-world applications, such as determining the optimal production levels in economics, analyzing the flow of fluids in engineering, and predicting the behavior of stock prices in finance. They are also used in various mathematical models and equations in physics and other scientific fields.

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