Question

Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ h(t)=\frac{t+2}{t^{2}+5 t+6} $$

Short answer

The derivative of the function \(h(t) = (t+2) / (t^2 + 5t + 6)\) using the quotient rule of differentiation is \(h'(t) = (t^2 - 4t + 4) / (t^2 + 5t + 6)^2\).

Step-by-step solution

Identify the functions

First, identify the functions that comprise the quotient. Here, take \(u(t) = t + 2\) and \(v(t) = t^2 + 5t + 6\).

Compute the derivatives of the functions

Next, find the derivatives of these functions. The derivative of \(u(t)\) with respect to \(t\) (denoted \(du/dt\)) is \(1\), and the derivative of \(v(t)\) with respect to \(t\) (denoted \(dv/dt\)) is \(2t + 5\).

Apply the quotient rule

Now, apply the quotient rule. The derivative of \(h(t)\)is \(h'(t) = (v(t) \cdot du/dt - u(t) \cdot dv/dt) / v(t)^2\). Substituting the known values, we find \(h'(t) = ((t^2 + 5t + 6) \cdot 1 - (t + 2) \cdot (2t + 5)) / (t^2 + 5t + 6)^2\). Expanding and simplifying this, we get \(h'(t) = (t^2 - 4t + 4) / (t^2 + 5t + 6)^2\).

Key concepts

Quotient Rule

In calculus, when you're given a function that is the ratio of two other functions, like \( h(t) = \frac{u(t)}{v(t)} \), the quotient rule is a convenient way to find the derivative of that function. It states that the derivative of \( h(t) \) is given by \( h'(t) = \frac{v(t) \cdot du/dt - u(t) \cdot dv/dt}{v(t)^2} \). Here's how you can apply it, broken down:

• Identify your top function as \( u(t) \) and your bottom function as \( v(t) \).
• Differentiate both \( u(t) \) and \( v(t) \) separately with respect to \( t \).
• Multiply the derivative of \( u(t) \) by \( v(t) \) and the derivative of \( v(t) \) by \( u(t) \), then subtract the second product from the first.
• Place that result over the square of \( v(t) \) to get the final derivative.

Be mindful while simplifying the resulting expression as it's easy to make algebraic errors. And remember, use the quotient rule ONLY when you have a function divided by another function.

Differentiation Rules

Differentiation rules are formulas and techniques used to find the derivative of a function. Some essential rules include the power rule, product rule, chain rule, and, as mentioned earlier, the quotient rule.

Power Rule

The power rule applies to functions of the form \( f(x) = x^n \), and the derivative is \( f'(x) = n \cdot x^{n-1} \).

Product Rule

The product rule is used when you have two functions multiplied together and states that \( (u \cdot v)' = u' \cdot v + u \cdot v' \).

Chain Rule

The chain rule is utilized when you have a composition of functions, and it states that if \( h(x) = f(g(x)) \), then \( h'(x) = f'(g(x)) \cdot g'(x) \).
Knowing when and how to apply these rules makes finding derivatives much more straightforward.

Simplifying Derivatives

Simplifying derivatives often involves algebraic manipulation to make the expression more manageable and easier to interpret. After applying differentiation rules like the quotient rule, you might be left with complex fractions or multiple terms that can be combined or factored.
Consider common algebraic strategies to simplify your derivative:

• Combine like terms that have the same powers of \( t \).
• Factor out common factors from the numerator and denominator.
• Cancel out any terms that appear in both the numerator and denominator.

Take extra care when simplifying to avoid mistakes, as such errors can misrepresent the behavior of the derivative and, in turn, the original function.

Calculus

Calculus is a branch of mathematics that deals with rates of change (derivatives) and accumulation of quantities (integrals). Through the use of differentiation, calculus enables us to analyze the behavior of functions, such as understanding their rates of change at different points and optimizing real-world scenarios. The exercise provided explores one aspect of calculus: differentiation. By mastering the rules and techniques of differentiation, such as the quotient rule, students can solve complex problems involving rates of change. Calculus is a tool not only for mathematics but also for various disciplines such as physics, engineering, economics, and biology, where it helps in modeling and solving practical problems.