Question

Composite functions. (a) Given the functions \(f(x, y(u))=x^{2}+3 y^{2}\) and \(y(u)=5 u+3\), express \(d f\) in terms of changes \(d x\) and \(d u\). (b) What is \(\left(\frac{\partial f}{\partial u}\right)_{x, u=1} ?\)

Short answer

Part (a): \[ d f = 2x \text{d}x + 30y \text{d}u \].Part (b): \( \frac{\text{∂}f}{\text{∂}u} = 240 \) at \( x, u = 1 \).

Step-by-step solution

Understanding the problem

Given two functions, the objective is to express the total differential of the composite function in terms of changes in independent variables and to find a partial derivative.

Finding the total differential of f(x, y)

The total differential of a function of two variables, such as \[ df = \frac{\text{∂}f}{\text{∂}x} \text{d}x + \frac{\text{∂}f}{\text{∂}y} \text{d}y \].For the given function \( f(x, y) = x^2 + 3y^2 \), calculate the partial derivatives: \[ \frac{\text{∂}f}{\text{∂}x} = 2x \] and \[ \frac{\text{∂}f}{\text{∂}y} = 6y \].

Expressing y as a function of u

The function \( y(u) = 5u + 3 \). Thus, the differential \[ \text{d}y = \frac{\text{d}y}{\text{d}u} \text{d}u = 5 \text{d}u \].

Substituting dy in the total differential

Substitute \( \text{d}y = 5 \text{d}u \) from Step 3 into the total differential found in Step 2: \[ df = 2x \text{d}x + 6y \text{d}y \]This becomes:\[ df = 2x \text{d}x + 6y (5 \text{d}u) = 2x \text{d}x + 30y \text{d}u \].

Calculating the partial derivative with respect to u at x, u = 1

From Step 4, the first part of df that concerns changes in u is \(30y \text{d}u \). At u = 1, compute y: \[ y(1) = 5(1) + 3 = 8 \].Thus,\( \frac{\text{∂}f}{\text{∂}u} = 30y = 30(8) = 240 \).

Key concepts

Total Differential

The concept of the total differential can be complicated, but let's break it down into manageable pieces. When we have a function of two variables, like our function \(f(x, y) = x^2 + 3y^2\), the total differential tells us how changes in both variables \(x\) and \(y\) affect the overall change in the function. The total differential is given by: \[ df = \frac{∂f}{∂x} dx + \frac{∂f}{∂y} dy \] Here’s what each term means: \(df\) is the total differential, \(\frac{∂f}{∂x}\) is the partial derivative of \(f\) with respect to \(x\), multiplied by the change in \(x (dx)\), and \(\frac{∂f}{∂y}\) is the partial derivative of \(f\) with respect to \(y\), multiplied by the change in \(y (dy)\). For our function \(f(x, y) = x^2 + 3y^2\), the partial derivatives are: \[ \frac{∂f}{∂x} = 2x, \quad \frac{∂f}{∂y} = 6y \] Now, substituting these into our total differential equation: \[ df = 2x dx + 6y dy \] This tells us how \(f\) changes based on small changes in \(x\) and \(y\).

Partial Derivatives

Partial derivatives are a fundamental concept in calculus. They represent how a function changes as one of its variables changes, holding the other variables constant. For our function \(f(x, y) = x^2 + 3y^2\), the partial derivatives are: \[ \frac{∂f}{∂x} = 2x, \quad \frac{∂f}{∂y} = 6y \] These derivatives tell us the rate of change of \(f\) with respect to \(x\) and \(y\). When we want to express \f\ in terms of another variable, like \(u\), we need to use the chain rule (which we'll discuss next) to find \(\frac{∂f}{∂u}\). In our exercise, we have a second function \(y(u) = 5u + 3\), which means that \(y\) is dependent on \(u\). To find \(\frac{∂f}{∂u}\), we combine partial derivatives using the chain rule.

Chain Rule

The chain rule is a crucial tool for dealing with composite functions. It allows us to find the derivative of a function based on another variable. For a function \(f(x, y)\) where \(y\) is a function of \(u\), the chain rule helps us express the changes with respect to \(u\). In our case, we have \(y(u) = 5u + 3\). The total differential with respect to \(x\) and \(u\) becomes: \[ df = 2x dx + 6y dy \] Knowing that \(dy = \frac{dy}{du} du = 5 du\), we substitute this into the differential: \[ df = 2x dx + 6y (5 du) = 2x dx + 30y du \] Now, to find \( \frac{∂f}{∂u} \), we focus on the \(du\) component: \[ \frac{∂f}{∂u} = 30y \] Given \(u = 1\), we compute \(y(1) = 5(1) + 3 = 8\). Thus, \[ \frac{∂f}{∂u} = 30 \cdot 8 = 240 \] So the partial derivative of \(f\) with respect to \(u\) when \(x\) and \(u = 1\) is 240. The chain rule elegantly ties together the changes in both \(x\) and \(u\) to show their combined impact on \(f\).