Problem 30 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the derivative of each function. \(h(t)=t^{2}(3 t+4)^{3}\)

Step-by-Step Solution

Answer

The short answer is: The derivative of the given function is \(h'(t) = (3t+4)^2(2t(3t+4) + 9t^2)\).

Step 1: Identify the functions to apply the Product Rule

Let's identify the two functions that we need to differentiate individually: \(u(t)=t^2\) \(v(t)=(3t+4)^3\) We are given the function: \(h(t) = u(t)v(t) = t^2(3t+4)^3\) Now, let's find the derivative of each function \(u(t)\) and \(v(t)\).

Step 2: Differentiate u(t)

To find the derivative of \(u(t) = t^2\), use the power rule, which says that the derivative of \(t^n\) equals \(n \cdot t^{n-1}\) -- in this case, \(n=2\): \(u'(t) = 2 \cdot t^{2-1} = 2t\)

Step 3: Differentiate v(t) using the Chain Rule

Here, we need to apply the chain rule to differentiate the function \(v(t)=(3t+4)^3\). To do that, first take the derivative of the outer function and evaluate it by keeping the inner function intact: \(\frac{d}{dt}[(3t+4)^3] = 3(3t+4)^{3-1}\) Then, multiply this result by the derivative of the inner function: \(\frac{d}{dt}[3t+4]=3\) Finally, multiply both derivatives together: \(v'(t) = 3(3t+4)^2 \cdot 3\)

Step 4: Apply the Product Rule

Now we have both the derivatives, \(u'(t)\) and \(v'(t)\), so it's time to apply the product rule to find the derivative of \(h(t)\). The product rule states that the derivative of a product of two functions equals the derivative of the first function times the second function plus the first function times the derivative of the second function: \(h'(t) = u'(t)v(t) + u(t)v'(t)\)

Step 5: Calculate the derivative h'(t)

Substitute the derivatives and functions we found earlier to obtain the derivative of h(t): \(h'(t) = (2t)(3t+4)^3 + (t^2) \cdot (9(3t+4)^2)\)

Step 6: Simplify the expression

We can simplify the expression by factoring out the common term \((3t+4)^2\): \(h'(t) = (3t+4)^2(2t(3t+4) + 9t^2)\) And there you have the derivative of the given function: \(h'(t) = (3t+4)^2(2t(3t+4) + 9t^2)\)

Key Concepts

Product Rule

Differentiation in calculus often requires us to find the derivative of functions that are products of other functions. The Product Rule is a handy tool for this. Imagine you have two functions, say, \(u(t)\) and \(v(t)\). Their product is expressed as \(h(t) = u(t) \times v(t)\). Unlike simple functions, the derivative of a product is not just the product of the derivatives. This is where the Product Rule shines.The rule states that you must take the derivative of the first function, multiply it by the second function, and then add the product of the first function and the derivative of the second function:

• First, find \(u'(t)\), the derivative of \(u(t)\).
• Next, find \(v'(t)\), the derivative of \(v(t)\).
• Apply the rule: \(h'(t) = u'(t)v(t) + u(t)v'(t)\).

This rule ensures you get every interaction the functions have in their multiplication, making your results accurate and comprehensive. It's a crucial concept when dealing with products in calculus.

Power Rule

The Power Rule is a fundamental concept in differentiation, primarily used for handling expressions with powers. It makes finding derivatives of polynomial functions extremely straightforward. The rule is simple. If you have a function \(u(t) = t^n\), the derivative, \(u'(t)\), can be quickly found using the formula:

• Bring down the exponent: The derivative is \(n\), the original exponent, times \(t\) raised to the power of \(n-1\).
• The rule mathematically is expressed as: \(u'(t) = n \cdot t^{n-1}\).

For example, to differentiate \(t^2\), apply the Power Rule to get \(2 \cdot t^{2-1} = 2t\).This rule simplifies the process significantly, allowing for quick computations and handling of terms that appear in any polynomial function or elsewhere. Mastering the Power Rule is essential for anyone working with calculus and algebra.

Chain Rule

The Chain Rule in calculus is essential for differentiating composite functions, where one function is nested inside another. This rule allows us to differentiate such complex structures efficiently. Here’s how it works when given a function like \(v(t) = (3t+4)^3\), where a function of \(t\) (\(3t+4\) in this case) is raised to a power.Applying the Chain Rule requires two steps:

• First, differentiate the outer function while treating the inner function as a constant. For \((3t+4)^3\), the outer function derivative becomes \(3(3t+4)^{3-1}\).
• Next, multiply this result by the derivative of the inner function. The derivative of \(3t+4\) is 3, so you'll end up with \(3(3t+4)^2 \cdot 3\).

By following these steps, you can effectively unravel the complexity of composite functions. The Chain Rule is a powerful technique to have in your calculus toolkit, especially when dealing with nested or composed expressions. It empowers you to handle a range of problems involving layers of functions.