Problem 110 - Calculus and its applications

Question

Find dy/dx. Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$ y=(x+1)^{3} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 3(x+1)^2 \)

Step 1: Identify the Function

The given function is a polynomial raised to a power: \( y = (x+1)^3 \). This function can be differentiated using the chain rule because it involves a composition of functions.

Step 2: Apply the Chain Rule

To differentiate \( y = (u)^3 \) where \( u = x+1 \), use the chain rule. The chain rule states that if a function \( y = f(u) \) and \( u = g(x) \), then \( \frac{dy}{dx} = f'(u)\cdot g'(x) \). Here, \( f(u) = u^3 \) and \( g(x) = x+1 \), so we need to find both \( f'(u) \) and \( g'(x) \).

Step 3: Differentiate the Outer Function

Differentiate the outer function \( y = u^3 \) with respect to \( u \). The derivative is \( f'(u) = 3u^2 \).

Step 4: Differentiate the Inner Function

Differentiating the inner function \( u = x+1 \) with respect to \( x \), we get \( g'(x) = 1 \).

Step 5: Apply the Chain Rule Formula

Substitute the derivatives from Step 3 and Step 4 into the chain rule formula: \( \frac{dy}{dx} = f'(u)\cdot g'(x) = 3u^2 \cdot 1 = 3u^2 \).

Step 6: Substitute Back the Inner Function

Replace \( u \) with \( x+1 \) to express \( \frac{dy}{dx} \) in terms of \( x \): \( \frac{dy}{dx} = 3(x+1)^2 \).

Key Concepts

Chain Rule in Differentiation

Understanding how to differentiate a function that is a composition of other functions is made simpler with the chain rule. The chain rule is essential when dealing with functions nested inside each other. Imagine peeling an onion; you work from the outer layer to the inner parts. In calculus, if you have a function like \( y = (x+1)^3 \), it appears as one smooth mathematical expression, but it's actually a composition of simpler functions. This is where the chain rule applies.

• First, identify the outer function and the inner function, which here are \( f(u) = u^3 \) and \( g(x) = x+1 \) respectively.
• The chain rule formula, \( \frac{dy}{dx} = f'(u) \cdot g'(x) \), helps guide how each piece contributes to the derivative.

Essentially, you take the derivative of the outer function with respect to the inner one and then multiply by the derivative of the inner function. This layered approach captures how small changes in \( x \) affect changes in \( y \) through the chain of functions.

Understanding Polynomial Functions

Polynomial functions form one of the simplest types of functions in calculus, yet they are pivotal for understanding how more complex functions behave. A polynomial function like \((x + 1)^3\) is simply a sum of terms consisting of variables raised to whole number exponents. They are smooth and continuous, making them easy to work with.

• A polynomial of degree 3, such as \( (x + 1)^3 \), means the highest power of \( x \) in its expanded form is 3.
• Polynomials of degree \( n \) have \( n-1 \) turning points, which reflect where the function changes from increasing to decreasing, or vice-versa.
• They are infinitely differentiable, meaning you can keep taking derivatives.

In our scenario, though we are working initially with \( (x + 1)^3 \), the underlying structure remains a polynomial, highlighting why rules like the chain rule become especially useful when working with such functions.

Derivative Rules for Calculus

Calculus offers a toolbox of rules that allow us to find derivatives quickly and efficiently. When differentiating functions, we often use derivative rules like the power rule, product rule, and chain rule, depending on the structure of the function.

• The power rule states: if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \). This is crucial for breaking down polynomial elements.
• The chain rule, previously discussed, helps differentiate compositions of functions.
• Combine the chain rule with the power rule for functions like \( (x+1)^3 \). Use the power rule on the outer function, \( u^3 \), then apply the chain rule to handle its composition with \( x+1 \).

These rules, applied in concert, allow you to tackle a wide variety of functions and their derivatives quickly. The derivative rules are foundational in calculus, serving as the building blocks for more advanced techniques and analyses in mathematics.