Problem 15 - Calculus for Biology and Medicine

Question

Use the product rule to find the derivative with respect to the independent variable. \(g(t)=3\left(2 t^{2}-5 t^{4}\right)^{2}\)

Step-by-Step Solution

Answer

The derivative of the function is \( g'(t) = 48t^3 - 300t^5 + 300t^7 \).

Step 1: Identify Outer and Inner Functions

The function given, \( g(t) = 3(2t^2 - 5t^4)^2 \), contains an outer function \( u(x) = x^2 \) and an inner function \( v(t) = 2t^2 - 5t^4 \). Recognizing these will help us apply the product and chain rule effectively.

Step 2: Apply the Chain Rule

To find the derivative of \( g(t) = 3 (v(t))^2 \), first, identify the derivative of the outer and inner functions. The chain rule states: \((f(g(t)))' = f'(g(t)) \cdot g'(t)\). Here, the derivative of \( u(x) = x^2 \) with respect to \( x \) is \( 2x \), so the derivative of \( u(v(t)) = (2t^2 - 5t^4)^2 \) is \( 2(2t^2 - 5t^4) \). Next, find \( v'(t) = \frac{d}{dt}(2t^2 - 5t^4) \).

Step 3: Differentiate the Inner Function

Differentiate \( v(t) = 2t^2 - 5t^4 \) with respect to \( t \). The derivative is \( v'(t) = \frac{d}{dt}(2t^2) - \frac{d}{dt}(5t^4) = 4t - 20t^3 \).

Step 4: Combine Derivatives Using the Chain Rule

Now apply the chain rule: \[ g'(t) = 3 \cdot 2(2t^2 - 5t^4) \cdot (4t - 20t^3) \]. Simplify this to get \[ g'(t) = 6(2t^2 - 5t^4)(4t - 20t^3) \].

Step 5: Simplify the Expression

Distribute to simplify: Expand \( 6(8t^3 - 40t^5 - 10t^5 + 50t^7) \). Simplifying, \[ g'(t) = 6(8t^3 - 50t^5 + 50t^7) \], which becomes \[ g'(t) = 48t^3 - 300t^5 + 300t^7 \].

Key Concepts

Chain Rule

The Chain Rule is a fundamental theorem in calculus that allows you to find the derivative of a composite function. A composite function is essentially a function that is made by combining two (or more) functions. To understand how the Chain Rule works, you need to break down the problem into its inner and outer functions.
For example, consider a function of the form \(h(x) = f(g(x))\). Here, \(g(x)\) is the inner function, and \(f(x)\) is the outer function.
The Chain Rule states:

• If \(h(x) = f(g(x))\), then the derivative \(h'(x)\) is \(f'(g(x)) \, \cdot \, g'(x)\).

Applying this rule involves:

• Differentiating the outer function while keeping the inner function unchanged, this is \(f'(g(x))\).
• Then multiplying by the derivative of the inner function, which is \(g'(x)\).

This process is essential when dealing with more complex functions like \(3(2t^2 - 5t^4)^2\), where identifying the inner and outer parts helps simplify what appears to be a complicated calculus problem.

Derivative

The Derivative is at the core of calculus, representing the rate at which a function is changing at any given point. Derivatives help us understand how a function behaves, be it moving up, down, or staying constant at a specific interval.
This concept is crucial in various fields, from physics to economics, where understanding change over time is essential.
Here's how you can think of derivatives more concretely:

• When you take the derivative of a function, you're finding the slope of the tangent line at any point on the graph of the function.
• It's a formal way of expressing "how one quantity changes with respect to another quantity."
• The notation for the derivative of a function \(f(x)\) is \(f'(x)\) or \(\frac{df}{dx}\).

In the exercise, finding the derivatives of the inner function \(2t^2 - 5t^4\) and the outer function \(x^2\) separately, then combining them using the Chain Rule helps in understanding the complete derivative of the given function by following these steps.

Calculus

Calculus is the branch of mathematics that deals with continuous change. It is divided into two main branches: differential calculus and integral calculus. Differential calculus concerns itself with understanding rates of change and slopes of curves, while integral calculus is about areas under or between curves.
In practical terms:

• Differential calculus allows for the calculation of gradients and changes, as seen in the process of finding derivatives.
• Integral calculus enables computation of quantities where a variable changes continuously, such as finding areas or volumes.

Understanding calculus requires a grasp of its fundamental rules and tools, such as the Chain Rule and the Product Rule.
These rules allow solving complex problems efficiently, as is required in the given exercise where the Product and Chain Rules are used in tandem to find the derivative of the function \(g(t) = 3(2t^2 - 5t^4)^2\).
Grasping these basic calculus concepts will not only help solve academic problems but also apply them in real-world scenarios. Utilizing calculus effectively provides insights into dynamic systems across natural and social sciences.