Question

$$\text { If } w=f(a x+b y), \text { show that } b \frac{\partial w}{\partial x}-a \frac{\partial w}{\partial y}=0$$. Hint: Let \(a x+b y=z\).

Short answer

\[ b \frac{\partial w}{\partial x} - a \frac{\partial w}{\partial y} = 0 \]

Step-by-step solution

- Substitution

Let’s use the hint provided. Define a new variable: \[ z = a x + b y \]Then the given function becomes: \[ w = f(z) \]

- Partial Derivative with respect to x

Use the chain rule to find \( \frac{\partial w}{\partial x} \). Since \( w \) depends on \( z \) and \( z \) depends on \( x \), we have:\[ \frac{\partial w}{\partial x} = \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial x} \]The partial derivative of \( z \) with respect to \( x \) is:\[ \frac{\partial z}{\partial x} = a \]Thus,\[ \frac{\partial w}{\partial x} = \frac{\partial w}{\partial z} \cdot a \]

- Partial Derivative with respect to y

Similarly, use the chain rule to find \( \frac{\partial w}{\partial y} \). Since \( w \) depends on \( z \) and \( z \) depends on \( y \), we have:\[ \frac{\partial w}{\partial y} = \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial y} \]The partial derivative of \( z \) with respect to \( y \) is:\[ \frac{\partial z}{\partial y} = b \]Thus,\[ \frac{\partial w}{\partial y} = \frac{\partial w}{\partial z} \cdot b \]

- Substitute back into the given equation

Substitute the expressions for \( \frac{\partial w}{\partial x} \) and \( \frac{\partial w}{\partial y} \) into the equation:\[ b \frac{\partial w}{\partial x} - a \frac{\partial w}{\partial y} \]Substitute:\[ b \left( \frac{\partial w}{\partial z} \cdot a \right) - a \left( \frac{\partial w}{\partial z} \cdot b \right) \]Simplify the terms:\[ b a \frac{\partial w}{\partial z} - a b \frac{\partial w}{\partial z} \]Which results in:\[ 0 \]

- Conclusion

Thus, we have shown that:\[ b \frac{\partial w}{\partial x} - a \frac{\partial w}{\partial y} = 0 \]

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus, especially when dealing with composite functions. It helps us take the derivative of a function based on its dependencies.
In our exercise, we used the chain rule to differentiate the function \( w = f(z) \) with respect to \( x \) and \( y \). First, let's recap the idea: if you have a composite function \( w = f(z) \) where \( z = g(x) \), the chain rule tells us that:
\[ \frac{\frac{\text{d} w}{\text{d} x}}{\text{d} x} = \frac{\text{d} w}{\text{d} z} \times \frac{\text{d} z}{\text{d} x} \text{ or in partial derivatives, } \frac{\frac{\text{w}}{\text{x}}}{\text{x}} = \frac{\text{w}}{\text{z}} \times \frac{\text{z}}{\text{x}}. \text{\begin{equation*}}
\]
This is what we applied when differentiating \( w \) with respect to \( x \) and \( y \). It's important because it breaks down the process into simpler steps. Remember, the chain rule allows us to propagate dependencies all the way through nested functions. This concept is especially useful in multivariable calculus and helps us get through complex differentiation problems.

Function of Multiple Variables

Many functions in real-life scenarios depend on more than one variable. For these functions, we talk about how they change with respect to each of their independent variables.
In mathematical terms, a function of multiple variables can be written as \( w = f(x, y) \). Here, \( w \) depends on both \( x \) and \( y \). Partial derivatives are used to explore how the function changes when one of the variables changes, keeping others constant.
In the context of our exercise, we initially had \( w = f(ax + by) \). To find how \( w \) changes with \( x \) and \( y \), we introduced the new variable \( z \), simplifying the function to \( w = f(z) \). This shifts our focus to how \( w \) changes as \( z \) changes, which in turn changes with respect to \( x \) and \( y \). Developing an understanding of these changes is key in fields like physics, engineering, and economics. So, knowing how to work with functions of multiple variables gives us powerful tools to tackle various real-world problems.

Implicit Differentiation

Implicit differentiation is necessary when you have equations where the variables are mixed together implicitly rather than given as an explicit function. In our original problem, we derived partial derivatives by substituting a new variable \( z \).
To explain further, consider an implicit equation like \( w = f(a x + b y) \). Here, \( w \) is not directly given as a function of \( x \) or \( y \), but rather through a combination of both. Implicit differentiation helps us extract meaningful derivative information without solving explicitly for one variable in terms of the others.
We introduced \( z = ax + by \) and then used this new variable to differentiate implicitly with respect to both \( x \) and \( y \). This made things straightforward by leveraging the chain rule:
\[ \frac{\text{w}}{\text{x}} = \frac{\text{w}}{\text{z}} \times \frac{\text{z}}{\text{x}} \text{ and } \frac{\text{w}}{\text{y}} = \frac{\text{w}}{\text{z}} \times \frac{\text{z}}{\text{y}}. \text{\begin{equation*}}\]
By understanding implicit differentiation, you'll be able to solve problems where variables are deeply intertwined. This is a skill that enhances your problem-solving toolbox.