Simple Partial Differentiation problem

In summary, the conversation is about proving a mathematical equation using partial derivatives. The equation is given and the person is trying to solve it step by step, showing their attempts and asking for help in case of any mistakes. The summary provides a step-by-step explanation of the solution and concludes that there doesn't seem to be any mistake in the solution.
  • #1
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Homework Statement


If ##z=x\ln(x+r)-r## where ##r^2=x^2+y^2##, prove that
$$\frac{∂^2z}{∂x^2}+\frac{∂^2z}{∂y^2}=\frac{1}{x+y}$$

Homework Equations


The Attempt at a Solution


Since ##r^2=x^2+y^2##, ##∂r/∂x=x/r## and ##∂r/∂y=y/r##.
Differentiating z w.r.t x partially,
$$\frac{∂z}{∂x}=\ln(x+r)+x\cdot\left(\frac{1}{x+r}\right)\cdot \left(1+\frac{∂r}{∂x}\right)-\frac{∂r}{∂x}$$
Using ##∂r/∂x=x/r##
$$\frac{∂z}{∂x}=\ln(x+r)$$
$$\frac{∂^2z}{∂x^2}=\frac{1}{x+r}\cdot\left(1+\frac{x}{r}\right)=\frac{1}{r}$$

Differentiating z w.r.t y partially,
$$\frac{∂z}{∂y}=\frac{x}{x+r}\cdot \frac{∂r}{∂y}-\frac{∂r}{∂y}=\frac{∂r}{∂y}\cdot \frac{-r}{x+r}$$
Using ##∂r/∂y=y/r##
$$\frac{∂z}{∂y}=\frac{-y}{x+r}$$
$$\frac{∂^2z}{∂y^2}=-\frac{(x+r)-y(∂r/∂y)}{(x+r)^2}=-\frac{rx+r^2-y^2}{r(x+r)^2}$$
Since ##r^2-y^2=x^2##
$$\frac{∂^2z}{∂y^2}=-\frac{x}{r(x+r)}$$
Adding the second order derivatives,
$$\frac{∂^2z}{∂x^2}+\frac{∂^2z}{∂y^2}=\frac{1}{r}-\frac{x}{r(x+r)}=\frac{1}{x+r}$$
Where did I go wrong? :confused:

Any help is appreciated. Thanks!
 
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  • #2
Pranav-Arora said:

Homework Statement


If ##z=x\ln(x+r)-r## where ##r^2=x^2+y^2##, prove that
$$\frac{∂^2z}{∂x^2}+\frac{∂^2z}{∂y^2}=\frac{1}{x+y}$$


Homework Equations





The Attempt at a Solution


Since ##r^2=x^2+y^2##, ##∂r/∂x=x/r## and ##∂r/∂y=y/r##.
Differentiating z w.r.t x partially,
$$\frac{∂z}{∂x}=\ln(x+r)+x\cdot\left(\frac{1}{x+r}\right)\cdot \left(1+\frac{∂r}{∂x}\right)-\frac{∂r}{∂x}$$
Using ##∂r/∂x=x/r##
$$\frac{∂z}{∂x}=\ln(x+r)$$
$$\frac{∂^2z}{∂x^2}=\frac{1}{x+r}\cdot\left(1+\frac{x}{r}\right)=\frac{1}{r}$$

Differentiating z w.r.t y partially,
$$\frac{∂z}{∂y}=\frac{x}{x+r}\cdot \frac{∂r}{∂y}-\frac{∂r}{∂y}=\frac{∂r}{∂y}\cdot \frac{-r}{x+r}$$
Using ##∂r/∂y=y/r##
$$\frac{∂z}{∂y}=\frac{-y}{x+r}$$
$$\frac{∂^2z}{∂y^2}=-\frac{(x+r)-y(∂r/∂y)}{(x+r)^2}=-\frac{rx+r^2-y^2}{r(x+r)^2}$$
Since ##r^2-y^2=x^2##
$$\frac{∂^2z}{∂y^2}=-\frac{x}{r(x+r)}$$
Adding the second order derivatives,
$$\frac{∂^2z}{∂x^2}+\frac{∂^2z}{∂y^2}=\frac{1}{r}-\frac{x}{r(x+r)}=\frac{1}{x+r}$$
Where did I go wrong? :confused:

I don't think you have; I also get the same answer.
 
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  • #3
pasmith said:
I don't think you have; I also get the same answer.

Thanks for the check pasmith! :smile:
 

1. What is a simple partial differentiation problem?

A simple partial differentiation problem involves finding the partial derivative of a function with respect to one of its variables, while holding all other variables constant. This allows us to measure the rate of change of a function in a specific direction.

2. How is simple partial differentiation different from regular differentiation?

Regular differentiation involves finding the derivative of a function with respect to one variable, while simple partial differentiation involves finding the partial derivative with respect to one variable while holding all other variables constant.

3. Why is simple partial differentiation important?

Simple partial differentiation is important in many fields of science, such as physics, economics, and engineering. It allows us to analyze how a function changes in a specific direction, which is useful in understanding real-world phenomena and making predictions.

4. What are the steps to solving a simple partial differentiation problem?

The steps to solving a simple partial differentiation problem are: 1) Identify the variable you are differentiating with respect to, 2) Treat all other variables as constants, 3) Use the rules of differentiation (such as the power rule or product rule) to find the partial derivative, and 4) Simplify the result.

5. Can you provide an example of a simple partial differentiation problem?

Yes, for example, let's say we have the function f(x,y) = 2x^2y + 3xy^2. To find the partial derivative with respect to x, we would treat y as a constant and use the power rule to get fx(x,y) = 4xy + 3y^2. Similarly, to find the partial derivative with respect to y, we would treat x as a constant and use the power rule to get fy(x,y) = 2x^2 + 6xy.

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