Problem 16 - The Calculus with Analytic Geometry

Question

Find the derivative of the given function. $$ f(y)=(y+3)^{3}(5 y+1)^{2}\left(3 y^{2}-4\right) $$

Step-by-Step Solution

Answer

The derivative is \(f'(y) = 3(y + 3)^2(5y + 1)^2(3y^2 - 4) + 10(y + 3)^3(5y + 1)(3y^2 - 4) + 6(y + 3)^3(5y + 1)^2 y\).

Step 1 - Identify the Product Rule

The function is a product of three functions: 1) \(u = (y + 3)^3\), 2) \(v = (5y + 1)^2\), and 3) \(w = (3y^2 - 4)\).We will apply the product rule: d[uv]=u'v+uv' and so on for three functions.

Step 2 - Find the Derivative of Each Function

Calculate the derivatives individually: For \(u = (y + 3)^3\): \(u' = 3(y + 3)^2\)For \(v = (5y + 1)^2\): \(v' = 2(5y + 1) \times 5 = 10(5y + 1)\)For \(w = (3y^2 - 4)\): \(w' = 6y\)

Step 3 - Apply the Product Rule

Apply the product rule for three functions: \( f'(y) = u'vw + uv'w + uvw' \)Substituting the calculated derivatives, \( f'(y) = 3(y + 3)^2 \times (5y + 1)^2 \times (3y^2 - 4) + (y + 3)^3 \times 10(5y + 1) \times (3y^2 - 4) + (y + 3)^3 \times (5y + 1)^2 \times 6y \)

Step 4 - Simplify the Expression

Combine and simplify the expression from the previous step.\( f'(y) = 3(y + 3)^2(5y + 1)^2(3y^2 - 4) + 10(y + 3)^3(5y + 1)(3y^2 - 4) + 6(y + 3)^3(5y + 1)^2 y \)

Key Concepts

product rule

When dealing with the derivative of a product of functions, it is essential to understand the product rule. The product rule states that if you have two functions, say \( u \) and \( v \), the derivative of their product \( uv \) is given by:
\[ (uv)' = u'v + uv' \]
This means you take the derivative of the first function and multiply it by the second function, and then add the product of the first function and the derivative of the second function.
In cases like our exercise where we have more than two functions, we extend this rule. If \( u, v, \) and \( w \) are three functions, then the derivative is:
\[ (uvw)' = u'vw + uv'w + uvw' \]
This principle allows us to tackle complicated expressions by breaking them down into simpler parts.

derivative calculation

Now, let's dive into actually calculating the derivatives of each component function. For our exercise, we had three functions:

• \( u(y) = (y + 3)^3 \)
• \( v(y) = (5y + 1)^2 \)
• \( w(y) = (3y^2 - 4) \)

First, compute the derivative of \( u(y) = (y + 3)^3 \). Using the power rule, we know:
\[ u'(y) = 3(y + 3)^2 \]
Second, compute the derivative of \( v(y) = (5y + 1)^2 \). Applying the chain rule, we get:
\[ v'(y) = 2(5y + 1) \times 5 = 10(5y + 1) \]
Lastly, for the derivative of \( w(y) = (3y^2 - 4) \), again using the power rule:
\[ w'(y) = 6y \]
Each derivative utilizes either the power rule or chain rule, simplifying our work efficiently.

chain rule in derivatives

The chain rule is another critical tool for finding derivatives, especially when functions are composed of other functions. The chain rule states:
\[ \frac{d}{dx} [f(g(x))] = f'(g(x)) \times g'(x) \]
In other words, to find the derivative of a composite function, you first take the derivative of the outer function, then multiply it by the derivative of the inner function.

For our current exercise, we used the chain rule in the process of finding the derivative of \( v(y) = (5y + 1)^2 \):
The outer function is \( g(z) = z^2 \) and the inner function is \( h(y) = 5y + 1 \). Thus,
\[ v'(y) = 2(5y + 1) \times 5 = 10(5y + 1) \]
Mastering the chain rule enriches your ability to handle complex derivative problems involving nested functions, making the derivative calculations easier.