Problem 89 - University Calculus: Early Transcendentals

Question

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=(x+1)^{x}$$

Step-by-Step Solution

Answer

The derivative is \( y' = (x+1)^x (\ln(x+1) + \frac{x}{x+1}) \).

Step 1: Take the Natural Logarithm of Both Sides

Start by taking the natural logarithm of both sides of the given function. This helps simplify the expression using the properties of logarithms. We have: \[ \ln(y) = \ln((x+1)^x) \] Apply the power rule for logarithms: \( \ln(a^b) = b \ln(a) \), giving us: \[ \ln(y) = x \ln(x+1) \]

Step 2: Differentiate Both Sides with Respect to x

Differentiate both sides of the equation with respect to \(x\). On the left side, use implicit differentiation. The derivative of the left side: \[ \frac{d}{dx} [\ln(y)] = \frac{1}{y} y' \] The derivative of the right side uses the product rule, which states that \(d [uv]/dx = u'v + uv'\): - \(u = x\), so \(u' = 1\)- \(v = \ln(x+1)\), so \(v' = \frac{1}{x+1}\)Therefore: \[ \frac{d}{dx} [x \ln(x+1)] = 1 \cdot \ln(x+1) + x \cdot \frac{1}{x+1} = \ln(x+1) + \frac{x}{x+1} \] Now, putting it all together: \[ \frac{1}{y} y' = \ln(x+1) + \frac{x}{x+1} \]

Step 3: Solve for y'

To find \(y'\), multiply both sides by \(y\). \[ y' = y \left(\ln(x+1) + \frac{x}{x+1}\right) \] Since we know \(y = (x+1)^x\), substitute back: \[ y' = (x+1)^x \left(\ln(x+1) + \frac{x}{x+1}\right) \]

Step 4: Simplify the Expression

Simplify the expression if possible. The derivative remains as is: \[ y' = (x+1)^x \left(\ln(x+1) + \frac{x}{x+1}\right) \] This is the derivative of \(y\) with respect to \(x\).

Key Concepts

Derivative

In mathematics, the derivative represents the rate at which a function is changing at any given point and is a fundamental tool in calculus. When performing derivatives, we essentially measure how a function's output changes as its input shifts. This is particularly useful when we need to find the slope of a curve at a point or determine the velocity of an object at a particular instant.

• The process of finding a derivative is called differentiation.
• The derivative of a function like \(f(x)\) with respect to \(x\) is denoted by \(f'(x)\) or \(\frac{df}{dx}\).
• It helps us understand the behavior of functions, like finding maximum or minimum points where the function's rate of change is zero.

Different techniques such as the product rule, chain rule, and implicit differentiation, are often used to find derivatives. In the exercise above, we used differentiation along with logarithms to simplify and solve more complex functions like \(y=(x+1)^x\). Logarithmic differentiation is used when functions involve variables in both the base and the exponent, allowing us to easily manage these more intricate cases.

Natural Logarithm

The natural logarithm, commonly denoted as \(\ln(x)\), is a logarithm with the base of the mathematical constant \(e\), where \(e\approx2.718\). It is used to transform multiplicative relationships into additive ones, making it a very powerful tool for simplifying expressions in calculus.

• One of the critical properties of natural logarithms is that \(\ln(e) = 1\).
• The natural logarithm helps in scaling inputs so that relations remain linearized, which makes differentiations manageable.

Using the natural logarithm allowed us to take the log of both sides of the equation, transforming \(y=(x+1)^x\) into \(\ln(y)=x\ln(x+1)\). This transformation was crucial to enable simpler differentiation of the function by applying the properties of logarithms, such as \(\ln(a^b) = b \ln(a)\).
Natural logarithms are widely used in various branches of science and engineering to deal with exponential growth models, decay processes, and to simplify mathematical models for better understanding.

Product Rule

The product rule is a technique used in calculus for differentiating products of two or more functions. It allows us to take the derivative of the product rather than dealing with each component separately. The general formula for the product rule when faced with two functions \(u(x)\) and \(v(x)\) is:\[\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)\]
This rule is especially handy when you have a function that can be neatly split into two parts:

• The derivative of the first function is found and then multiplied by the second function unchanged.
• The first function is taken unchanged and multiplied by the derivative of the second function.

These results are then added together to give the derivative of the whole product. For example, in the original exercise, \(x \ln(x+1)\) required the product rule to differentiate:
- \(u = x\) and \(u' = 1\)- \(v = \ln(x+1)\) and \(v' = \frac{1}{x+1}\)Using the product rule, the derivative became \(\ln(x+1) + \frac{x}{x+1}\). This method helps in situations where two differentiable functions are multiplied together, facilitating complex derivative calculations, especially in logarithmic differentiation.