Problem 3 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find \(g^{\prime}(t)\) (a) \(g(t)=\frac{2^{t}}{t^{2 t}}\), where \(t>0\) (b) \(g(t)=\ln (t+1)^{t^{2}+1}\), where \(t>-1\)

Step-by-Step Solution

Answer

The solutions for the derivatives are \(g^{\prime}(t) = \ln(2) \cdot 2^{t} \cdot t^{-2t} - 2^{t} \cdot t^{2t - 1} \cdot (1 + 2t \ln(t))\) for the first function, and \(g^{\prime}(t) = \frac{(t^{2}+1) + (t+1) \cdot 2t}{t+1}\) for the second function.

Step 1: Find the Derivative of \(g(t)=\frac{2^{t}}{t^{2 t}}\)

\[g^{\prime}(t) = u^{'}v + uv^{'}\] where \[u = 2^{t}\] and \[v = \frac{1}{t^{2t}}\]Applying the chain rule, we can determine \(u^{'}\) as \[u^{'} = 2^{t} \ln(2)\])For \(v^{'}\), we must apply both the derivative of a quotient rule, yielding\[v^{'} = -t^{2t - 1} (1 + 2t \ln(t))\]Therefore our answer is given by \[g^{\prime}(t) = \ln(2) \cdot 2^{t} \cdot t^{-2t} - 2^{t} \cdot t^{2t - 1} \cdot (1 + 2t \ln(t))\]

Step 2: Find the Derivative of \(g(t)=\ln ((t+1)^{t^{2}+1})\)

\[g^{\prime}(t) = \frac{1}{t+1} \cdot d((t+1)^{t^{2}+1})\]The inner function \[d(u) = u^{'} \cdot (t^{2}+1) + u \cdot (2t)\] where \(u = t+1\)We evaluate \(u^{'}\) first to get \[u^{'} = 1\]So substituting this we get \[d(u) = (t^{2}+1) + (t+1) \cdot (2t)\]Therefore our answer is \[g^{\prime}(t) = \frac{(t^{2}+1) + (t+1) \cdot 2t}{t+1}\]

Key Concepts

Chain Rule

Understanding the chain rule is essential when dealing with complex functions where one function is composed within another. Essentially, the chain rule assists in finding the derivative of a composite function by taking the derivative of the outer function and multiplying it by the derivative of the inner function.

For example, consider a function written as \(h(x) = f(g(x))\), where \(f\) and \(g\) are both functions of \(x\). To find \(h'(x)\), the derivative of \(h\) with respect to \(x\), the chain rule tells us it is \(f'(g(x))\) times \(g'(x)\). From a practical standpoint, the steps are as follows:

• Identify the outer function (\(f\)) and the inner function (\(g\)).
• Compute the derivative of the outer function (\(f'(g(x))\)) with respect to the inner function.
• Compute the derivative of the inner function (\(g'(x)\)) with respect to \(x\).
• Multiply these derivatives together to obtain the final result.

Applying this to the mentioned exercise, when finding the derivative of \(g(t) = 2^t / t^{2t}\), the chain rule is crucial in obtaining \(u'\) for the numerator part \(2^t\), simplifying the process to \( u' = 2^t \cdot \ln(2)\).

Quotient Rule

The quotient rule is a procedure used to differentiate functions that are divided by each other—ratios of functions. When one function is divided by another, \(\frac{f(x)}{g(x)}\), the quotient rule explains how to find the derivative \(\left( \frac{f}{g} \right)'\) by taking \(f'(x)g(x) - f(x)g'(x)\) all over \(g(x)^2\).

The steps for the quotient rule are:

• Find the derivatives of the numerator \(f'(x)\) and the denominator \(g'(x)\) separately.
• Apply the formula \(\frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\).

In our exercise, to determine \(v'\), the derivative of the denominator \(v = \frac{1}{t^{2t}}\), we combine the quotient rule and the chain rule. This results in \(v' = -t^{2t - 1} (1 + 2t \ln(t))\). The quotient rule is especially important because it addresses the division of two functions, which is a common occurrence in calculus.

Logarithmic Differentiation

Logarithmic differentiation is a technique used to simplify the process of finding the derivatives of functions that can be challenging to differentiate using standard rules. This method is particularly useful when dealing with functions raised to the power of another function or the products and quotients of functions that are not easily differentiated by straightforward application of the basic rules.

The general approach to logarithmic differentiation is to:

• Take the natural logarithm (\(\ln\)) of both sides of the equation \(y = f(x)\).
• Utilize properties of logarithms to simplify the expression.
• Differentiate implicitly with respect to \(x\).
• Solve for \(\frac{dy}{dx}\) or \(y'\).

For instance, part (b) of the exercise presents the function \(g(t) = \ln ((t+1)^{t^2+1})\). To find its derivative, we apply logarithmic differentiation, which helps us break down the expression into more manageable parts and is particularly advantageous when the exponent (\(t^2+1\)) depends on \(t\). The process leads to a simpler form, allowing us to differentiate and find \(g'(t)\) without directly applying more complex rules to \( (t+1)^{t^2+1} \).