Problem 63 - Calculus an Applied Approach

Question

Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ y=t^{2} \sqrt{t-2} $$

Step-by-Step Solution

Answer

The derivative of the function is \( y' = 2t\sqrt{t-2} + \frac{1}{2}t\sqrt{t} \). The product rule and the power rule for differentiation were used to find this derivative.

Step 1: Identify the functions in the product

Firstly, identify the two functions that are being multiplied together in the given function. This can be done by recognizing that the function \( y = t^{2} \sqrt{t-2} \) is a product of two functions \( f(t) = t^{2} \) and \( g(t) = \sqrt{t-2} \).

Step 2: Convert the function to be suitable for differentiation

To make it easier to differentiate, you need to express \( \sqrt{t-2} \) as \( (t-2)^{0.5} \) and the original equation becomes: \( y = t^{2}(t-2)^{0.5} \).

Step 3: Apply the Product Rule of differentiation with the Power Rule

Using the product rule of differentiation (u'v + uv') and then subsequently applying the power rule for finding derivatives, the derivative of the original function is calculated as: \( y' = d/dt(t^{2}(t-2)^{0.5}) = 2t(t-2)^{0.5} + 0.5t^{2}(t-2)^{-0.5}(1) \).

Step 4: Simplification

Simplify the above expression to get the final expression: \( y' = 2t\sqrt{t-2} + \frac{1}{2}t\sqrt{t} \).

Key Concepts

Product Rule in Differentiation

The product rule is a fundamental principle used in calculus when taking the derivative of a product of two functions. This is essential because, in calculus, we often encounter functions that are products of simpler functions. The product rule states that if you have two differentiable functions, say \( f(t) \) and \( g(t) \), then the derivative of their product \( y = f(t) g(t) \) can be calculated as:

• \( y' = f'(t)g(t) + f(t)g'(t) \)

To put this simply, you take the derivative of the first function and multiply it by the second function, then add the product of the second function's derivative and the first function. This rule helps break down complex problems into simpler parts, making it critical for solving derivatives of functions like \( y = t^{2} \sqrt{t-2} \), where each function has its specific derivative rules.

Power Rule for Derivatives

The power rule is another cornerstone of differentiation. It provides a quick and straightforward way to find the derivative of functions expressed in the form of \( t^n \). This rule is crucial when dealing with terms of a function that have an exponent. The power rule states:

• If \( y = t^n \), then \( y' = n t^{n-1} \).

This means that you bring down the exponent as a coefficient in front, then reduce the exponent by one. For example, the derivative of \( t^2 \) is \( 2t^{1} = 2t \). This simplicity makes the power rule an essential tool, especially when combined with the product rule in differentiating complex expressions like \( y = t^{2}(t-2)^{0.5} \). When dealing with roots, remember to convert them into fractions: \( \sqrt{t-2} \) becomes \( (t-2)^{0.5} \).

Understanding Differentiation

Differentiation is the fundamental process in calculus used to calculate the rate at which a function is changing at any point. This is analogous to finding the slope of the tangent line to the function at a given point. For any function \( y = f(t) \), differentiation yields \( y' = \frac{dy}{dt} \), which is the derivative of the function. Differentiation involves techniques and rules to handle various types of functions, such as polynomial, exponential, trigonometric, and more complex combinations.
- **Finding Derivatives**: It involves applying basic rules like the power rule and product rule.
- **Simplifying the Process**: By breaking down the problem into more manageable parts, such as differentiating products or sums individually and then combining them using appropriate rules.
This concept allows us to understand how a function behaves, how fast it changes, and enables predictions of trends and patterns, vital in scientific and engineering applications. Thus, mastering differentiation is crucial for analyzing and interpreting the behavior of functions in various contexts.