Problem 20 - Single Variable Calculus

Question

Find the derivative of the function. \(h(t)=\frac{t^{5}-3 t^{3}+2 t^{2} e^{t}}{2 t^{2}}\)

Step-by-Step Solution

Answer

The derivative of the function \(h(t) = \frac{t^{5}-3 t^{3}+2 t^{2} e^{t}}{2 t^{2}}\) is: \(h'(t) = 2e^t\)

Step 1: Find the derivatives of the numerator and denominator functions.

First, let's find the derivatives of both the numerator and denominator functions separately. We will need them to apply the quotient rule. \(g(t) = t^{5} - 3t^{3} + 2t^{2} e^{t}\), so: \(g'(t) = 5t^{4} - 9t^{2} + 4t e^t + 2t^{2} e^t\) Now, for the denominator: \(h(t) = 2t^{2}\), so: \(h'(t) = 4t\)

Step 2: Apply the quotient rule to find the derivative of the function

Now that we have the derivatives of both the numerator and denominator functions, we can apply the quotient rule to find the derivative of the function h(t): \[h'(t) = \frac{g'(t) \cdot h(t) - g(t) \cdot h'(t)}{[h(t)]^{2}}\] Substituting the expressions we found in Step 1: \(h'(t) = \frac{(5t^{4} - 9t^{2} + 4t e^t + 2t^{2} e^t) \cdot (2t^{2}) - (t^{5}-3 t^{3}+2 t^{2} e^{t}) \cdot (4t)}{[2t^{2}]^{2}}\)

Step 3: Simplify the expression

Now let's simplify the expression for the derivative: \(h'(t) = \frac{10t^{6} - 18t^{4} + 8t^{3} e^t + 4t^{4} e^t - 4t^{6} + 12t^{4} - 8t^{3} e^t}{4t^{4}}\) Combine the like terms: \(h'(t) = \frac{(10t^{6} - 4t^{6}) + (-18t^{4} + 12t^{4} + 4t^{4} e^t)}{4t^{4}}\) Simplify: \(h'(t) = \frac{6t^{6} - 6t^{4} + 4t^{4} e^t}{4t^{4}}\) Now, we can factor out a \(2t^{4}\) from the expression in the numerator: \(h'(t) = \frac{2t^{4}(3 - 3 + 2 e^t)}{4t^{4}}\) Finally, we can simplify the expression by dividing the numerator and denominator by \(2t^{4}\): \(h'(t) = 3 - 3 + 2e^t\) So, the derivative of the function h(t) is: \(h'(t) = 2e^t\)

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental method in calculus for finding the derivative of a division of two functions. If you have a function that is the quotient of two functions, say \( h(t) = \frac{g(t)}{d(t)} \), then the derivative \( h'(t) \) can be found with the formula:

• \( h'(t) = \frac{g'(t) \cdot d(t) - g(t) \cdot d'(t)}{[d(t)]^2} \)

Using this formula correctly requires that you first find the derivatives of the numerator (\( g'(t) \)) and the denominator (\( d'(t) \)) separately.
It’s crucial to maintain the structure and sign of the formula, as the subtraction in the numerator is essential for the correct outcome.
Applying it allows you to differentiate complex rational expressions step by step.

Derivative of Exponential Functions

Exponential functions, like \( e^t \), have a special characteristic in differentiation.
The derivative of such a function remains an exponential function. Specifically, the derivative of \( e^t \) is simply \( e^t \) itself.
When using the chain rule with exponential functions, if the exponent includes a function of \( t \), then it must also be differentiated:

• For \( e^{f(t)} \), the derivative is \( f'(t) \cdot e^{f(t)} \).
• This is because of the chain rule, which accounts for the rate of change "inside" the exponential function.

Including exponential derivatives, as in our example, requires careful attention.
Exponential derivatives are implemented when we apply calculus rules to find how quickly things change under exponential growth.

Simplifying Algebraic Expressions

Simplifying algebraic expressions in calculus is all about making the equation easier to read and solve.
Often after applying calculus operations, you'll end up with unwieldy expressions packed full of terms. The goal is to combine like terms and factor where possible.

• Combining like terms means you add or subtract coefficients of terms that have the exact same variable powers.
• Factoring is another powerful simplification tool, useful for pulling common elements out of terms to make them easier to work with.

After obtaining the derivative for our example, simplifying brought us down from a complex fraction to a much simpler form.
This process not only clears the path for further calculus operations but also serves to highlight the true behavior of the function's changes.