Problem 62 - Calculus an Applied Approach

Question

Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ y=t \sqrt{t+1} $$

Step-by-Step Solution

Answer

The derivative of the function \(y = t \sqrt{t+1}\) is \(y' = \sqrt{t+1} + \frac{t}{2\sqrt{t+1}}\). The Product Rule and the Chain Rule of differentiation were used in the problem.

Step 1: Using the Product Rule

The Product Rule states that the derivative of two multiplied functions is the derivative of the first function multiplied by the second function as it is, plus the first function as it is multiplied by the derivative of the second function. Here, we first let one function be \(t\) and the other be \(\sqrt{t+1}\). So, using the Product Rule, the derivative \(y'\) becomes: \(y' = 1 * \sqrt{t+1} + t * (1/2) * (t+1)^{-1/2} * 1\)

Step 2: Simplifying the Expression

Now the expression needs to be simplified to its simplest form:\(y' = \sqrt{t+1} + \frac{t}{2\sqrt{t+1}}\)

Step 3: Applying the Chain Rule

Remember that the expression \( (t+1)^{-1/2} \) was derived using the Chain Rule. The Chain Rule is applied to differentiate a function composed of an outer function and an inner function. In this case, the outer function is \( (x)^{-1/2} \) and the inner function is \( (t + 1) \), which when differentiated gives 1.

Key Concepts

Product Rule

The Product Rule is a key differentiation technique in calculus. When you have two functions multiplied together, and you want to find the derivative of their product, that's when the Product Rule shines. Imagine you have a function \( y = f(t) \cdot g(t) \), where both \( f(t) \) and \( g(t) \) are functions of \( t \). The Product Rule helps you find the derivative of \( y \), denoted as \( y' \).

The formula for the Product Rule is:

• \( y' = f'(t) \cdot g(t) + f(t) \cdot g'(t) \)

In simple terms, you:

• Take the derivative of the first function \( f(t) \), multiply it with the second function \( g(t) \) as it is.
• Then, take the original first function \( f(t) \) as it is, and multiply it with the derivative of the second function \( g(t) \).

Add them up, and you have the derivative of the product. This rule was applied in our original problem where \( t \) and \( \sqrt{t+1} \) were considered as two separate functions.

Chain Rule

The Chain Rule is essential when dealing with composite functions, where one function is nested inside another. If you have a function inside of another function, and you need to differentiate it, the Chain Rule is your best friend. Let's break it down.

Suppose you have a function \( g(f(t)) \), where \( f(t) \) is an inner function and \( g(u) \) is an outer function, with \( u = f(t) \). To find the derivative of \( g(f(t)) \), follow these steps:

• First, differentiate the outer function \( g(u) \) with respect to \( u \).
• Then, multiply the result by the derivative of the inner function \( f(t) \) with respect to \( t \).

This gives you the derivative of a composite function.

In our problem, the expression \( (t+1)^{-1/2} \) was a result of applying the Chain Rule. Here, \( (t+1) \) was the inner function and \( u^{-1/2} \) was the outer function.

Differentiation Techniques

Differentiation is the process of finding the derivative, which tells us how a function changes. It brings us essential tools like the Product Rule and Chain Rule to tackle complex expressions and functions.

Differentiation techniques provide us with the framework to:

• Break down intricate problems with ease.
• Find out how functions behave and predict future trends or changes.
• Create simplified expressions and make calculations manageable.

For the problem at hand, we used a combination of these powerful rules. Knowing when to apply which rule is crucial. The Product Rule was useful because there was a multiplication of functions. Meanwhile, the Chain Rule was necessary to handle functions that were nested, as seen with \( \sqrt{t+1} \).

Understanding these techniques not only helps in finding derivatives but also builds the foundation for the rest of calculus, allowing you to solve variable rates of change in scientific, engineering, and mathematical problems.