Problem 26 - College Algebra and Calculus: An Applied Approach

Question

Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ h(t)=\left(t^{5}-1\right)\left(4 t^{2}-7 t-3\right) $$

Step-by-Step Solution

Answer

The derivative of the given function is \( h'(t) = 28t^6 - 35t^5 - 8t^2 - 7 \).

Step 1 - Identify the Functions

Identify the two functions that are being multiplied. We can see that function1, \(f(t)\), is \(t^5 - 1\) while function2, \(g(t)\), is \(4t^2 - 7t - 3\).

Step 2 - Find the Derivatives of \(f(t)\) and \(g(t)\)

Now find the derivatives of \(f(t)\) and \(g(t)\). The derivative \(f'(t)\) is \(5t^4\), derived using the power rule of differentiation. The derivative \(g'(t)\) is \(8t - 7\), also derived using the power rule.

Step 3 - Apply the Product Rule

According to the product rule of differentiation, the derivative of \( h(t) = f(t)g(t) \) is given by \( h'(t) = f(t)g'(t) + g(t)f'(t) \). Therefore, by substituting the functions and their derivatives from Step 1 and Step 2, we get \( h'(t) = (t^5-1)(8t - 7) + (4t^2 -7t -3)(5t^4) \).

Step 4 - Simplify the Expression

Simplify the equation according to the above equation to get the final answer. After multiplication and simplification, the result is \( h'(t) = 20t^6 - 35t^5 + 8t^6 - 7t - 8t^2 + 7 \). Combine like terms to give the final answer, \( h'(t) = 28t^6 - 35t^5 - 8t^2 - 7 \).

Key Concepts

Power Rule

The power rule is a straightforward method used in differentiation. It's handy for finding derivatives of functions that are polynomials. Basically, the power rule states that if you have a function in the form of \(x^n\), where \(n\) is any real number, the derivative of this function is given by \(nx^{n-1}\).

• For example, if your function is \(t^5\), its derivative would be \(5t^4\).
• This is because you multiply the exponent (5) by the coefficient (1), and then subtract one from the exponent to get 4.

The power rule is widely used because many functions are polynomials or contain polynomial terms. So whenever you see a term like \(t^5\) or \(8t\) in a function, that's your cue to make use of the power rule in differentiation.

Product Rule

The product rule is essential when you need to differentiate a product of two functions. If you have two functions, say \(f(t)\) and \(g(t)\), and they are multiplied together in a function \(h(t) = f(t)g(t)\), the product rule guides you to find the derivative of \(h(t)\). The rule tells us that the derivative is ordered as follows: take the derivative of the first function, multiply it by the second function unchanged, and then add the product of the first function unchanged times the derivative of the second function. This is mathematically represented as:\[ h'(t) = f'(t)g(t) + f(t)g'(t) \]

• In our original problem, \(f(t) = t^5 - 1\) and \(g(t) = 4t^2 - 7t - 3\).
• To find \(h'(t)\), we calculate \(f'(t) = 5t^4\) and \(g'(t) = 8t - 7\) using the power rule, then apply them in the product rule formula.

Thus, it combines both the unchanged and the differentiated parts of each function, giving us a comprehensive way to derive functions that are products of other functions.

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative represents the rate at which a function is changing at any given point. When you're tasked to differentiate a function, you're usually looking to find this rate of change or the slope of the function at any point on its curve. Many mathematical rules and methods, like the power rule and product rule, help us compute derivatives efficiently. Differentiation has practical applications in various fields:

• In physics, it's used to calculate velocity and acceleration.
• In economics, derivatives can help find profit maximization and cost minimization.
• In everyday life, it can help us understand anything that changes, like speed, growth, or decline over time.

In the original exercise, differentiation helped us find the derivative \(h'(t)\), allowing us to understand exactly how the function \( h(t) = (t^5-1)(4t^2-7t-3) \) changes with respect to \( t \). By solidly using the principles of differentiation, concepts within calculus become clearer and calculable.