Problem 18 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the derivative of each function. \(f(t)=\frac{1-2 t}{1+3 t}\)

Step-by-Step Solution

Answer

The derivative of the function \(f(t)=\frac{1-2t}{1+3t}\) is \(f'(t) =\frac{-5}{(1 + 3t)^2}\).

Step 1: Identify u(t) and v(t)

In the function \(f(t)=\frac{1-2t}{1+3t}\), we can identify: \(u(t)= 1-2t\) \(v(t)= 1+3t\) Now, we can find the derivatives of \(u(t)\) and \(v(t)\).

Step 2: Find the derivatives of u(t) and v(t)

In order to calculate the derivatives, recall that \((cf)'(t) = c\cdot f'(t)\) for any constant \(c\) and function \(f\). Then we have: \(u'(t) = \frac{d}{dt}(1 - 2t) = 0 - 2 = -2\) \(v'(t) = \frac{d}{dt}(1 + 3t) = 0 + 3 = 3\)

Step 3: Apply the quotient rule

Now that we have found \(u(t)\), \(v(t)\), their derivatives, we can apply the quotient rule to find \(f'(t)\): \(f'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2}\) Plug our expressions for \(u(t)\), \(v(t)\), \(u'(t)\), and \(v'(t)\): \(f'(t) = \frac{(-2)(1 + 3t) - (1 - 2t)(3)}{(1 + 3t)^2}\)

Step 4: Simplify the expression

By simplifying the given expression, we will be able to find the derivative of the function: \(f'(t) =\frac{-2 - 6t - 3 + 6t}{(1 + 3t)^2}\) Simplifying further: \(f'(t) =\frac{-5}{(1 + 3t)^2}\) The derivative of the given function, \(f(t)\), is \(f'(t) =\frac{-5}{(1 + 3t)^2}\).

Key Concepts

Quotient Rule

The quotient rule is a vital tool in calculus used for finding the derivative of a division between two functions. When you have a function expressed as a fraction, such as \(f(t) = \frac{u(t)}{v(t)}\), the quotient rule applies. It's designed to handle these fractions and ensure the derivative is computed correctly. For the quotient rule, the formula is:\[f'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2}\]This formula might seem a bit intimidating at first, but it's quite straightforward once you break it down step by step.

• Numerator: You calculate \(u'(t)\) (the derivative of the numerator part) and \(v(t)\) (leave the denominator part as it is), then subtract the product of \(u(t)\) (the original numerator) and \(v'(t)\) (the derivative of the denominator part).
• Denominator: You just square the denominator \([v(t)]^2\).

When using the quotient rule, it's important to clearly identify which function is \(u\) (the top function) and which is \(v\) (the bottom function) before finding their derivatives. This clarity ensures accurate application of the rule, as we saw in the exercise where \(u(t) = 1-2t\) and \(v(t) = 1+3t\) were used.

Differentiation

Differentiation is a core concept in calculus that involves finding the rate at which a function is changing at any given point. When we differentiate a function, we're essentially looking to uncover its derivative, which represents an instant rate of change.To differentiate a function like \(f(t) = 1 - 2t\), we focus on applying basic rules such as the power rule or constant rule. For the linear function here, the differentiation process is straightforward:

• For \(1\), a constant, the derivative is 0.
• For \(-2t\), using the constant multiple rule where the derivative of a constant times a function is constant times the derivative of the function. So, \(\frac{d}{dt}(-2t) = -2\).

Knowing these basic rules helps you find the derivative quickly and easily, especially with linear functions which are common in introductory calculus problems like the one given. By systematically applying these differentiation rules, one masterfully simplifies expressions to proceed with further calculations, such as using the quotient rule or other calculus methods.

Function Simplification

Simplifying functions is crucial after applying calculus rules to ease the interpretation of derivatives or results. Simplification involves reducing complex fractions or expressions into more manageable forms.In the given exercise, after applying the quotient rule, we obtained the expression:\[f'(t) = \frac{-2 - 6t - 3 + 6t}{(1 + 3t)^2}\]The steps in function simplification include:

• Combining like terms: In the numerator, terms \(-6t\) and \(+6t\) cancel each other, and constants \(-2 - 3\) sum up to \(-5\).
• Result: This yields a much simpler form, \(f'(t) = \frac{-5}{(1 + 3t)^2}\).
• Final Interpretation: This simplified derivative \(\frac{-5}{(1 + 3t)^2}\) is easier to work with and plug into further analyses or graphs to discern behavior of \(f(t)\).

Simplifying functions not only helps in reducing computation errors but also makes derivatives or any mathematical result easy to interpret or apply to various problems. It’s the final polish in the calculus process.