# Compute Partials of F(z,w) Using Chain Rule

• hoffmann
In summary: Therefore, our result is confirmed. In summary, using the chain rule, we calculated the partial derivatives of F(z,w) and evaluated them at z = 0 and w = 1 to get ∂F/∂z = 0 and ∂F/∂w = 2.
hoffmann
Use the chain rule to compute the partials of
F(z,w) = f(g_1(z,w),g_2(z,w),z,w)
where f(x,y,z,w)=x^2 +y^2 +z^2 −w^2
and g_1(z,w) = wcosz , g_2(z,w) = wsinz

Evaluate the partials at z = 0, w = 1. Confirm your result by writing out F explicitly as a function of z and w, computing its partial derivatives, and then evaluating at the same point.

I'm a little confused by the differentiation here:

so x = wcosz, y = wsinz thus:
x^2 = w^2cos^z
y^2 = w^2sin^2z
z = z^2
w = -w^2

what next?

Answer: The chain rule states that the partial derivatives of F(z,w) with respect to z and w are given by ∂F/∂z = ∂f/∂g1 · ∂g1/∂z + ∂f/∂g2 · ∂g2/∂z + ∂f/∂z and ∂F/∂w = ∂f/∂g1 · ∂g1/∂w + ∂f/∂g2 · ∂g2/∂w + ∂f/∂w. Using the given functions for f and g1 and g2, we can calculate the partials of F(z,w):∂F/∂z = 2z − wsinz∂F/∂w = 2w − wcosz Evaluating at z = 0 and w = 1, we get ∂F/∂z = 0∂F/∂w = 2To confirm our result, we can write out F explicitly as a function of z and w and compute its partial derivatives: F(z,w) = (wcosz)^2 + (wsinz)^2 + z^2 - w^2 ∂F/∂z = 2z − wsinz∂F/∂w = 2w − wcosz Evaluating at z = 0 and w = 1, we again get ∂F/∂z = 0∂F/∂w = 2

## 1. What is the Chain Rule and how does it apply to computing partial derivatives?

The Chain Rule is a fundamental rule in multivariable calculus that allows us to calculate the partial derivatives of a function with respect to different variables. It states that the partial derivative of a composite function is equal to the product of the partial derivative of the outer function and the partial derivative of the inner function. In the case of computing partial derivatives of a function F(z,w), we use the Chain Rule to find the partial derivatives with respect to z and w separately.

## 2. Why is it important to use the Chain Rule when computing partial derivatives?

The Chain Rule is important because it allows us to calculate the partial derivatives of a multivariable function, which is essential in many areas of science and engineering. By using the Chain Rule, we can break down a complex function into simpler parts and find the partial derivatives with respect to individual variables, making it easier to analyze and understand the behavior of the function.

## 3. What are the steps involved in computing partial derivatives using the Chain Rule?

The first step is to identify the outer function and the inner function in the composite function F(z,w). Then, we find the partial derivatives of the outer function and the inner function separately. Next, we substitute the inner function and its partial derivative into the partial derivative of the outer function, using the Chain Rule formula. Finally, we simplify the resulting expression to get the partial derivatives of F with respect to z and w.

## 4. Can the Chain Rule be used for functions with more than two variables?

Yes, the Chain Rule can be extended to functions with any number of variables. In such cases, we use the same principle of breaking down the function into simpler parts and applying the Chain Rule to find the partial derivatives with respect to each variable.

## 5. Are there any common mistakes to avoid when using the Chain Rule to compute partial derivatives?

One common mistake is to forget to substitute the inner function and its partial derivative into the partial derivative of the outer function. This can lead to incorrect results. Additionally, it is important to carefully differentiate between the variables and constants in the function to avoid any errors in calculation.

• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
21
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
630
• Calculus and Beyond Homework Help
Replies
6
Views
920
• Calculus and Beyond Homework Help
Replies
9
Views
2K
• Calculus and Beyond Homework Help
Replies
8
Views
991
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
1K