Question

(a) Use the Quotient Rule to differentiate the function. (b) Manipulate the function algebraically and differentiate without the Quotient Rule. (c) Show that the answers from \((\mathrm{a})\) and \((\mathrm{b})\) are equivalent. $$f(t)=\frac{t^{2}-1}{t+1}$$

Short answer

The derivative of \(f(t)\) is 1 using both methods.

Step-by-step solution

Apply the Quotient Rule

To find the derivative of \(f(t)=\frac{t^{2}-1}{t+1}\) using the Quotient Rule, remember that the Quotient Rule states if \(h(t) = \frac{u(t)}{v(t)}\), then \(h'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2}\). Here, \(u(t) = t^2 - 1\) and \(v(t) = t+1\). Compute \(u'(t) = 2t\) and \(v'(t) = 1\). Substituting these into the Quotient Rule formula gives us:

Simplify the Quotient Rule Derivative

Start with: \(f'(t) = \frac{(2t)(t+1) - (t^2 - 1)(1)}{(t+1)^2}\). Simplify the numerator: \((2t)(t+1) = 2t^2 + 2t\) and \((t^2 - 1)(1) = t^2 - 1\). Thus, the numerator becomes \(2t^2 + 2t - t^2 + 1 = t^2 + 2t + 1\). Therefore, \(f'(t) = \frac{t^2 + 2t + 1}{(t+1)^2}\).

Algebraic Manipulation

Rewrite \(f(t) = \frac{t^2-1}{t+1}\) by dividing each term of the numerator by the denominator: \(t^2 - 1 = (t^2 + t - t - 1) = (t+1)(t-1)\). Hence, \(f(t) = t - 1 + \frac{0}{t+1} = t - 1\) since \(\frac{0}{t+1} = 0\) for \(t eq -1\).

Differentiate the Simplified Function

The function is \(f(t) = t - 1\). Differentiating this with respect to \(t\) gives \(f'(t) = 1\).

Confirm Equivalence of Both Methods

From Step 2, we originally got \(f'(t) = \frac{t^2 + 2t + 1}{(t+1)^2}\). Factor the numerator as a perfect square: \((t+1)^2\), so \(f'(t) = \frac{(t+1)^2}{(t+1)^2} = 1\) when \(t eq -1\). This confirms the results from Steps 4 and 1 yield the same derivative.

Key concepts

Derivative

In calculus, the derivative of a function quantifies how the function's output changes as its input changes. It is essentially the "rate of change" or "slope" of the function at any given point. When dealing with functions presented as ratios, specifically, the Quotion Rule is often applied. The rule is a handy tool developed to determine the derivative of functions written as one function divided by another.

To apply this rule, you must understand both the function on the top (numerator) and the one on the bottom (denominator). These are traditionally represented as \( u(t) \) and \( v(t) \), respectively. The rule then states: if \( h(t) = \frac{u(t)}{v(t)} \), then the derivative \( h'(t) \) is given by this formula:

• \( h'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2} \)

This provides a straightforward process to locate the derivative of a quotient, necessitating systematic differentiation of both the numerator and the denominator. Yet, always remember to simplify thoroughly for a clean result.

Function Differentiation

Function differentiation involves taking the derivative of each component of a composite function. In the case of the problem given, we differentiate \( f(t) = \frac{t^2-1}{t+1} \) by applying the Quotient Rule to manage the more complex composition of the functions in its numerator and denominator.

First, focus on finding the derivatives for the individual components involved:

• Let \( u(t) = t^2 - 1 \), then \( u'(t) = 2t \).
• Let \( v(t) = t + 1 \), then \( v'(t) = 1 \).

Applying these into the Quotient Rule formula involves plugging these derivatives back into the rule, performing multiplications, and handling subtractions as prescribed by the rule. During this, careful algebraic manipulation ensures the result reflects the correct rate of change for the original function accurately.

Beyond procedural understanding, recognize when simplifications derive from perfect square factors or other identities, as this is crucial in verifying equivalence at different stages of solving.

Algebraic Manipulation

Algebraic manipulation is a powerful technique to simplify expressions, making further calculus steps significantly easier. In some cases, functions represented as complicated ratios can be rewritten to reveal simpler forms, as seen in our problem example.

The function \( f(t) = \frac{t^2-1}{t+1} \) was initially rewritten, recognizing that the numerator \( t^2 - 1 \) is a difference of squares: \( (t+1)(t-1) \). This realization divides completely into \( t+1 \), simplifying the expression to \( f(t) = t - 1 \).

By reformulating the function, it becomes straightforward to differentiate directly using basic differentiation rules instead of the Quotient Rule. This step marks how breaking down functions into simpler terms can yield not only an easier analysis but also an intuitive understanding, following identical derivative results for both steps indicate consistency of mathematical properties across different methods.