Finding the partial derivative from the given information

In summary, the conversation discusses the application of the multi-variable chain rule to a function with multiple inputs, specifically the function z = f(g(3r^3 - s^2, re^s)). The equation for the chain rule is provided, along with the values for the partial derivatives of the inputs. The conversation also addresses the need for the function f(g) in order to find the derivative of z with respect to g, and concludes with a clarification on how to use the given information to find the derivative of z with respect to g.
  • #1
Amadeo
28
9
Homework Statement
see post
Relevant Equations
dz/dr = dz/dx (dx/dr) + dz/dy(dy/dr)
problem13.PNG


It seems that the way to combine the information given is

z = f ( g ( (3r^3 - s^2), (re^s) ) )

we know that the multi-variable chain rule is

(dz/dr) = (dz/dx)* dx/dr + (dz/dy)*dy/dr

and

(dz/ds) = (dz/dx)* dx/ds + (dz/dy)*dy/ds

---(Parentheses indicate partial derivative)

other perhaps useful information

(dx/dr)= 9r^2
(dx/ds)=-2s
(dy/dr)=e^s
(dy/ds)=re^s

I don't know how to apply this information because, usually, there is only one variable, like t, being fed into the multi-input function, and the chain rule works nicely. But here we have r and s being fed into the multi-input function g. Further, g is the input of the function f, which is z. In order to obtain the derivative of z with respect to g, we would need to know the function f(g). It is not given. I would guess that the way to find the derivative of z with respect to r would be to multiply the derivative of z with respect to g by the derivative of g with respect to x, multiplied by the derivative of x with respect to r, plus the derivative of z with respect to g multiplied by the derivative of g with respect to y, multiplied by the derivative of y with respect to r.

So that is where I am on this. Thanks for any assistance.
 
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  • #2
Amadeo said:
here we have r and s being fed into the multi-input function g.
So apply the Relevant Equation you listed, with g in place of z.
Amadeo said:
g is the input of the function f, which is z. In order to obtain the derivative of z with respect to g, we would need to know the function f(g).
No, you only need to know the appropriate derivatives at the point of interest. You are given some values for gx, gy and f' (i.e., fg).
 
  • #3
Thank you. got it.

42, -24.
 

FAQ: Finding the partial derivative from the given information

What is a partial derivative?

A partial derivative is a mathematical concept used in calculus to describe the rate of change of a function with respect to one of its variables, while holding all other variables constant.

How do you find the partial derivative of a function?

To find the partial derivative of a function, you must first identify which variable you are taking the derivative with respect to. Then, you treat all other variables as constants and use the standard rules of differentiation to find the derivative.

What is the purpose of finding the partial derivative?

The purpose of finding the partial derivative is to understand how a function changes with respect to a specific variable, while keeping all other variables constant. This is useful in many fields, including physics, economics, and engineering, to analyze and optimize complex systems.

Can you give an example of finding a partial derivative?

Sure, let's say we have a function f(x,y) = 3x^2y + 5xy^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) = 6xy + 5y^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) = 3x^2 + 10xy.

What are some real-world applications of partial derivatives?

Partial derivatives have many applications in various fields, such as determining the optimal production levels in economics, analyzing the flow of fluids in engineering, and predicting the behavior of particles in physics. They are also used in machine learning and data analysis to optimize models and algorithms.

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