Question

If \(w=f(x, s, t), s=2 x+y, t=2 x-y,\) find \((\partial w / \partial x)_{y}\) in terms of \(f\) and its derivatives.

Short answer

\(\frac{\text{∂} w}{\text{∂} x} = \frac{\text{∂} f}{\text{∂} x} + 2 \frac{\text{∂} f}{\text{∂} s} + 2 \frac{\text{∂} f}{\text{∂} t}\)

Step-by-step solution

- Identify the given functions

Given functions are: \(w = f(x, s, t)\), \(s = 2x + y\), and \(t = 2x - y\).

- Express the partial derivatives with the chain rule

Using the chain rule for partial differentiation, we have: \(\frac{\text{d} w}{\text{d} x} = \frac{\text{∂} f}{\text{∂} x} + \frac{\text{∂} f}{\text{∂} s} \frac{\text{d} s}{\text{d} x} + \frac{\text{∂} f}{\text{∂} t} \frac{\text{d} t}{\text{d} x}\).

- Compute the derivatives of intermediate variables

Calculate the partial derivatives of \(s\) and \(t\) with respect to \(x\): \(\frac{\text{d} s}{\text{d} x} = 2\) and \(\frac{\text{d} t}{\text{d} x} = 2\).

- Substitute the derivatives into the chain rule equation

Substitute the derived values into the chain rule equation: \(\frac{\text{d} w}{\text{d} x} = \frac{\text{∂} f}{\text{∂} x} + \frac{\text{∂} f}{\text{∂} s} \times 2 + \frac{\text{∂} f}{\text{∂} t} \times 2\).

- Simplify the expression

Combine the terms to get the final expression: \(\frac{\text{d} w}{\text{d} x} = \frac{\text{∂} f}{\text{∂} x} + 2 \frac{\text{∂} f}{\text{∂} s} + 2 \frac{\text{∂} f}{\text{∂} t}\).

Key concepts

chain rule in calculus

Understanding the chain rule is crucial in calculus, especially for dealing with complex functions. The chain rule allows us to differentiate composite functions. For functions of a single variable, if we have a composition of functions such as \(g(h(x))\), the chain rule tells us: \[ \frac{\text{d} g}{\text{d} x} = \frac{\text{d} g}{\text{d} h} \times \frac{\text{d} h}{\text{d} x} \].

When functions have multiple variables, the same principle applies but we must consider partial derivatives. For example, if \(w=f(x, s, t)\) with intermediate variables defined as \(s=2x+y\) and \(t=2x-y\), we use the partial derivative chain rule:

\[ \frac{\text{d} w}{\text{d} x} = \frac{\text{∂} f}{\text{∂} x} + \frac{\text{∂} f}{\text{∂} s} \frac{\text{d} s}{\text{d} x} + \frac{\text{∂} f}{\text{∂} t} \frac{\text{d} t}{\text{d} x} \].

This rule is very useful when calculating derivatives of functions dependent on other variable functions.

multivariable functions

Multivariable functions involve functions that depend on more than one variable. For instance, let’s consider \(w = f(x, s, t)\). Here, \(w\) is dependent on three variables: \(x\), \(s\), and \(t\).

These functions often arise in real-world scenarios, such as physics or economics, where a phenomenon might depend on several factors.

To differentiate such functions, we need to recognize each variable's influence and use partial derivatives. When we take the partial derivative of \(w\) with respect to \(x\) while keeping \(y\) constant, we treat \(s\) and \(t\) as dependent on \(x\): \[s=2x+y\] and \[t=2x-y\].

This leads us to the use of chain rule to correctly compute the rate of change of \(w\) with respect to \(x\).

intermediate variables

In calculus, intermediate variables act as connecting points between the main variables and the function of interest. For example, in the exercise given, we are provided with intermediate variables \(s = 2x + y\) and \(t = 2x - y\), which depend on the main variables \(x\) and \(y\).

These intermediate variables help us break down the complex derivative calculations into simpler steps, allowing us to use the chain rule effectively.

When computing the partial derivative of \(w=f(x, s, t)\) with respect to \(x\), it is crucial to differentiate these intermediate variables with respect to \(x\) as well. This step-by-step process makes our calculations manageable and ensures accuracy.

For instance, calculating how \(s\) and \(t\) change with \(x\) is straightforward and gives us: \[ \frac{\text{d} s}{\text{d} x} = 2 \text{ and } \frac{\text{d} t}{\text{d} x} = 2 \].

Incorporating these results back into the main partial derivative of \(w\) completes the use of the chain rule to find \[ \frac{\text{d} w}{\text{d} x} = \frac{\text{∂} f}{\text{∂} x} + 2 \frac{\text{∂} f}{\text{∂} s} + 2 \frac{\text{∂} f}{\text{∂} t} \].