Problem 46 - Brief Applied Calculus

Question

Find the derivative of each function by using the Quotient Rule. Simplify your answers. $$ f(t)=\frac{2 t^{2}+t-5}{t^{2}-t+2} $$

Step-by-Step Solution

Answer

The derivative is \( f'(t) = \frac{-3t^2 + 17t - 3}{(t^2 - t + 2)^2} \).

Step 1: Identify the Components of the Quotient Rule

The Quotient Rule states that for a function given by \( f(x) = \frac{g(x)}{h(x)} \), the derivative \( f'(x) \) is \( \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \). Here, identify \( g(t) = 2t^2 + t - 5 \) and \( h(t) = t^2 - t + 2 \).

Step 2: Find the Derivative of the Numerator, g(t)

Calculate \( g'(t) \), the derivative of the numerator \( g(t) = 2t^2 + t - 5 \). Using the power rule, \( g'(t) = 4t + 1 \).

Step 3: Find the Derivative of the Denominator, h(t)

Calculate \( h'(t) \), the derivative of the denominator \( h(t) = t^2 - t + 2 \). Using the power rule, \( h'(t) = 2t - 1 \).

Step 4: Apply the Quotient Rule

Substitute the derivatives into the Quotient Rule formula: \( \frac{g'(t) h(t) - g(t) h'(t)}{(h(t))^2} \). This gives:\[\frac{(4t + 1)(t^2 - t + 2) - (2t^2 + t - 5)(2t - 1)}{(t^2 - t + 2)^2}\]

Step 5: Simplify the Expression

Simplify the expression in the numerator, which involves distributing and combining like terms.\[(4t + 1)(t^2 - t + 2) = 4t^3 - 4t^2 + 8t + t^2 - t + 2 = 4t^3 - 3t^2 + 7t + 2\]\[(2t^2 + t - 5)(2t - 1) = 4t^3 - 2t^2 + 2t^2 + t - 10t + 5 = 4t^3 - 10t + 5\]Subtract these results: \[4t^3 - 3t^2 + 7t + 2 - (4t^3 - 10t + 5) = -3t^2 + 17t - 3\]Thus, the derivative is:\[\frac{-3t^2 + 17t - 3}{(t^2 - t + 2)^2}\]

Step 6: Final Simplified Derivative

Write the final simplified derivative:\[f'(t) = \frac{-3t^2 + 17t - 3}{(t^2 - t + 2)^2}\]

Key Concepts

Quotient Rule

When you have a function that's a fraction or ratio like \( f(t) = \frac{2t^2 + t - 5}{t^2 - t + 2} \), the Quotient Rule is your go-to method for finding its derivative. This rule is like a special tool for handling derivatives of these types of functions. It tells us how to find the derivative of a quotient (which is just a fancy word for division) of two functions.

Let's break the rule down: if you have a function \( \frac{g(t)}{h(t)} \), its derivative, using the Quotient Rule, will be \( \frac{g'(t)h(t) - g(t)h'(t)}{(h(t))^2} \). This means we need the derivatives of the top and bottom parts of the fraction separately.

In this context, \( g(t) \) is the numerator (top part), which is \( 2t^2 + t - 5 \), and \( h(t) \) is the denominator (bottom part), \( t^2 - t + 2 \). Finding these derivatives and plugging them into the formula is how the Quotient Rule helps us find the derivative of the entire fraction.

Simplifying Expressions

Once you've applied the Quotient Rule, you'll often end up with a pretty complex expression that needs simplifying. In our example, after applying the rule, we had:\[\frac{(4t + 1)(t^2 - t + 2) - (2t^2 + t - 5)(2t - 1)}{(t^2 - t + 2)^2}\]Simplifying means making this expression easier to work with. We do this by expanding the terms and combining any like terms.

For example, multiply each term in the first part \((4t + 1)(t^2 - t + 2)\) to get:

• \(4t^3 - 4t^2 + 8t\)
• and then combining with \(t^2 - t + 2\) gives you \(4t^3 - 3t^2 + 7t + 2\)

Do the same for the second part of the numerator. Then, subtract it and combine terms:

• \(4t^3 - 10t + 5\) from \(4t^3 - 3t^2 + 7t + 2\)
• This simplifies to \(-3t^2 + 17t - 3\)

Reducing expressions isn't just about reducing the size but putting everything into a form that makes further work or understanding easier.

Power Rule

The Power Rule is one of the foundational tools in calculus for finding derivatives. Whenever you see expressions with powers of \( t \), like \( 2t^2 \), this rule is incredibly useful.

The Power Rule states that the derivative of \( t^n \) is \( nt^{n-1} \). In simpler terms, you bring the power down in front of \( t \) and reduce the exponent by one.

For instance, applying the Power Rule to our numerator function \( g(t) = 2t^2 + t - 5 \) involves:

• For \(2t^2\), bring down the 2, resulting in \(4t^1\), which simplifies to \(4t\).
• For \(t = t^1\), you get \(1 \cdot t^{0} = 1\) since any variable to the power of zero is 1.
• The constant \(-5\) becomes 0 since constants have no risk, simplifying to \(g'(t) = 4t + 1\).

This simple rule is a lifesaver when attacked by polynomials, making derivatives doable even for complicated functions.