Question

Given \(w=a y^{2}\) and \(y=b x^{2}+c x,\) find \(d w / d x\) by the chain rule.

Short answer

\(\frac{dw}{dx} = 2a (b x^2 + c x) (2b x + c)\).

Step-by-step solution

Identify the Chain Rule

The chain rule is used to differentiate composite functions. It states that if you have a function composition where \(z = f(y)\) and \(y = g(x)\), then the derivative of \(z\) with respect to \(x\) is given by \(\frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx}\).

Express \(w\) as a Function of \(y\)

Given \(w = a y^2\), this is clearly a function of \(y\). Here, \(w\) depends on \(y\) as \(w(y) = a y^2\).

Differentiate \(w\) with Respect to \(y\)

Compute the derivative of \(w\) with respect to \(y\): \[ \frac{dw}{dy} = \frac{d}{dy}(a y^2) = 2a y \].

Express \(y\) as a Function of \(x\)

Given \(y = b x^2 + c x\), where \(y\) is clearly a function of \(x\).

Differentiate \(y\) with Respect to \(x\)

Compute the derivative of \(y\) with respect to \(x\): \[ \frac{dy}{dx} = \frac{d}{dx}(b x^2 + c x) = 2b x + c \].

Apply the Chain Rule to Find \(\frac{dw}{dx}\)

Use the chain rule to find \(\frac{dw}{dx}\) by multiplying \(\frac{dw}{dy}\) by \(\frac{dy}{dx}\): \[ \frac{dw}{dx} = \frac{dw}{dy} \cdot \frac{dy}{dx} = (2a y) \cdot (2b x + c) \].

Substitute \(y\) Back into the Equation

Since \(y = b x^2 + c x\), substitute \(y\) back into the expression for \(\frac{dw}{dx}\):\[ \frac{dw}{dx} = 2a (b x^2 + c x) (2b x + c) \].

Key concepts

Composite Functions

Composite functions are essentially functions built from other functions. Imagine a machine where you first perform one operation, then take the result, and use it as input for another operation. Mathematically, if you have two functions, say \(f(x)\) and \(g(x)\), a composite function using \(g(x)\) followed by \(f(x)\) is written as \(f(g(x))\).
In the context of the given exercise, we see a similar pattern. Here, \(w = a y^2\) acts as our outer function, and \(y = b x^2 + c x\) is the inner function. By substituting, we create a composition where you begin by calculating \(y\) from \(x\), and then use \(y\) to find \(w\). Understanding this composition is essential as it forms the backbone of applying the chain rule correctly.
Composite functions are invaluable as they allow us to simplify complex calculations by breaking them down into smaller, more manageable problems.

Differentiation

Differentiation is a crucial concept in calculus that involves finding the rate at which a function changes at any given point. It tells us about the slope of the function's graph or, in simpler terms, how steep the function is at a particular point.
In practical terms, differentiation is used to find derivatives. The process involves applying specific rules to functions so we can determine their rates of change. For example, if you have a function \(f(x) = x^2\), its derivative \(f'(x)\) is found using basic differentiation rules like the power rule, which gives us \(f'(x) = 2x\).
In the exercise, we differentiate two functions: first \(w = a y^2\) with respect to \(y\), and then \(y = b x^2 + c x\) with respect to \(x\). These steps enable us to use the chain rule to eventually find \(\frac{dw}{dx}\). Differentiation helps us translate physical problems into mathematical language, where we can solve them using calculus.

Derivative of Composite Functions

Understanding how to find the derivative of composite functions is an important skill in calculus. This typically involves the chain rule, which is a method for differentiating such functions.
When differentiating a composite function, you consider both the inner and outer functions. The goal is to calculate the derivative of the outer function with respect to the inner function, and then multiply it by the derivative of the inner function itself.
In the example given, \(w = a y^2\) is differentiated with respect to \(y\) to get \(\frac{dw}{dy} = 2a y\), and \(y = b x^2 + c x\) is differentiated with respect to \(x\) to get \(\frac{dy}{dx} = 2b x + c\). These derivatives are combined as per the chain rule to find the derivative of the overall function\(\frac{dw}{dx}\).

• Apply the derivative rules appropriately for each step.
• Ensure the derivatives of both inner and outer functions are calculated.
• Use substitution to express results in terms of the original variables.

This approach ensures accuracy when tackling complex functions.

Calculus

Calculus is a branch of mathematics that focuses on change and motion. It is divided into two main parts: differential calculus, which deals with rates of change, and integral calculus, which involves the accumulation of quantities.
In this exercise, we employed techniques from differential calculus, mainly the chain rule, which is vital for understanding how changes in one variable affect another.
Some key ideas in calculus are:

• Limits: These help us define the derivative and integrate functions.
• Continuity: A must for performing calculus correctly.
• Differentiability: Determines if the derivative can be calculated.

By grasping these essentials, students can better appreciate and solve calculus problems like differentiation, integration, and beyond. Everyone studying calculus should explore how these concepts relate to real-world scenarios, making learning engaging and practical.