Problem 19 - Calculus for Biology and Medicine

Question

In Problems 1-28, differentiate the functions with respect to the independent variable. $$ f(r)=\left(r^{2}-r\right)^{3}\left(r+3 r^{3}\right)^{-4} $$

Step-by-Step Solution

Answer

Differentiate using the product and chain rules to get a complex expression.

Step 1: Identify the Functions

Notice that the given function is a product of two functions: \( u(r) = (r^2 - r)^3 \) and \( v(r) = (r + 3r^3)^{-4} \). Let's differentiate this product using the product rule.

Step 2: Differentiate the First Function

Use the chain rule to differentiate \( u(r) = (r^2 - r)^3 \). The derivative \( u'(r) \) is:\[3(r^2 - r)^2 \cdot (2r - 1)\].

Step 3: Differentiate the Second Function

Use the chain rule to differentiate \( v(r) = (r + 3r^3)^{-4} \). The derivative \( v'(r) \) is:\[-4(r + 3r^3)^{-5} \cdot (1 + 9r^2)\].

Step 4: Apply the Product Rule

The product rule states \( (uv)' = u'v + uv' \). Apply this to our functions:\[f'(r) = (3(r^2 - r)^2 (2r - 1)) (r + 3r^3)^{-4} + (r^2 - r)^3 (-4(r + 3r^3)^{-5} (1 + 9r^2))\].

Step 5: Simplify the Expression

Expand and simplify the expression for \( f'(r) \). Begin by distributing and collecting like terms, though seeing as this can be rather complex, recognize that this process might lead to a large, unwieldy polynomial in its expanded form.

Key Concepts

Product Rule in Differentiation

When faced with the task of differentiating a product of two functions, the product rule is your best friend.
It helps us efficiently navigate through a situation where functions are multiplied together and need to be differentiated.
Here's what the product rule tells us: if you have two functions, say \( u(r) \) and \( v(r) \), then the derivative of their product is given by the formula:

• \((uv)' = u'v + uv'\)

This might look a bit overwhelming at first, but let's break it down:
You take the derivative of the first function (\( u' \)), multiply it by the second function, and then add it to the first function multiplied by the derivative of the second function.
It’s like combining their individual rates of change to get the overall rate of change of the entire product.
In our problem, \( u \) is the first chunk \((r^2 - r)^3 \) and \( v \) is the second chunk \((r + 3r^3)^{-4} \).
So apply the product rule following the steps we talked about, and you'll be on the right track!

Understanding the Chain Rule

The chain rule is wonderful when you have a function within another function, often called a composite function.
Think of it as peeling an onion, layer by layer.
You differentiate the outer layer first, and then the inner layer.
How does this work in practice? Let's dive in:

• The chain rule states: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\)

That reads as: find the derivative of the outer function, keeping the inner function as it is, and then multiply by the derivative of the inner function.
It's a systematic approach to dealing with these nested functions.
In our example, for \( u(r) = (r^2 - r)^3 \), \( (r^2 - r) \) is the inner part and needs to be tackled firstly by the chain rule.
Similarly, \( v(r) = (r + 3r^3)^{-4} \) involves using the chain rule where \( (r + 3r^3) \) is differentiated as the inner function.
With practice, mastering the chain rule becomes much smoother, and it's an invaluable tool for tackling complex differentiation tasks!

Fundamentals of Function Differentiation

Differentiation, in simple terms, is about finding how a function changes as its input changes.
It's like a zoom lens slowly revealing gradual changes over a curve.
At the core, we recognize that differentiation provides us with a new function, known as the derivative, which tells us the rate of change at any given point.
Each type of function has its differentiation method:

• Polynomials are straightforward – multiply by the exponent and reduce the exponent by one.
• For products and compositions, use the product rule and chain rule respectively, as mentioned earlier.
• When functions involve powers or roots, logarithmic or exponential rules may come in handy.

Differentiation acts as a powerful tool, whether in physics for velocity or in business for cost analysis.
Understanding the basics, like applying rules appropriately, helps solve complex problems seamlessly.