# Partial differential with respect to y

• issisoccer10
In summary, we are trying to find the partial derivative f_y(0,0) for the function f(x,y) = (x^3 + y^3)^(1/3). Using the chain rule, we get two possible solutions: 1/3(x^3 + y^3)^(-2/3) * 3y^2 or y^2 * (x^3 + y^3)^(-2/3). However, when plugging in (0,0), neither solution yields 1. This is because the function f is not differentiable at (0,0), although the partial derivatives exist. A more detailed analysis is required to find
issisoccer10
[SOLVED] Partial differential with respect to y

## Homework Statement

Given the equation f(x,y) = (x$$^{3}$$ + y$$^{3}$$)$$^{1/3}$$

Show that $$f_{y}$$(0,0) = 1

## Homework Equations

Basic chain rule..

## The Attempt at a Solution

Based on the chain rule...I believe that

$$f_{y}$$ = 1/3(x$$^{3}$$ + y$$^{3}$$)$$^{-2/3}$$ 3y$$^{2} or f_{y}$$ = y$$^{2}$$(x$$^{3}$$ + y$$^{3}$$)$$^{-2/3}$$

However, plugging in (0,0) does make $$f_{y}$$ equivalent to 1. Where is my mistake? Any help would be greatly appreciated.. thanks

The problem is that the function f, itself, is not differentiable at (0,0) even though the partial derivatives exist at (0,0). (But the partial derivatives are not continuous at (0,0).)

Go back to the original definition of the derivative:
At (0,0), f(x,y)= f(0,0)= 0.

For h not 0, f(0,0+h)= f(0,h)= ((h)3)1/3= h.
$$\frac{\partial f}{\partial y}(0,0)= \lim_{h\rightarrow 0}\frac{f(0,h)- f(0,0)}{h}= \lim_{h\rightarrow 0}\frac{h- 0}{h}= 1$$

Yah, you can tell right away something is fishy because the work you did, if you actually plug in x and y = 0, not only doesn't yield 1, it gives you 0/0

A bit of thinking and you'll see the original f(x,y) is undefined for x and y less than 0(one of them can be negative, but x^3+y^3 must be positive)

So at the point (0,0) the function kinda "comes into existence", so it's like a sharp sudden point where it begins, instead of being [math words that I forget], so you know you need to do a slightly more detailed analysis to find the derivative

## 1. What is a partial differential with respect to y?

A partial differential with respect to y is a type of mathematical derivative that measures the rate of change of a function with respect to the variable y, while keeping all other variables constant.

## 2. How is a partial differential with respect to y different from a regular derivative?

A regular derivative measures the rate of change of a function with respect to a single variable, while a partial differential with respect to y takes into account the effect of multiple variables on the function.

## 3. What is the purpose of taking a partial differential with respect to y?

A partial differential with respect to y is often used in multivariable calculus and physics to model and analyze systems with multiple variables. It allows us to understand how changes in one variable affect the overall system.

## 4. How do you calculate a partial differential with respect to y?

To calculate a partial differential with respect to y, you take the derivative of the function with respect to y while treating all other variables as constants. This is often done using the chain rule or product rule.

## 5. Can a partial differential with respect to y be negative?

Yes, a partial differential with respect to y can be negative. This indicates that the function is decreasing with respect to y at that point. However, the overall function can still be increasing if the effect of other variables outweighs the decrease in y.

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