Question

Consider the following functions (on the given interval, if specified). Find the derivative of the inverse function. $$f(x)=e^{3 x+1}$$

Short answer

Answer: The derivative of the inverse function of f(x) = e^(3x+1) is g'(x) = 1/(3x).

Step-by-step solution

Find the inverse function of f(x)

We have $$f(x) = e^{3x+1}$$. To find the inverse function, we can first call it $$g(x)$$, and express it as $$x = e^{3g(x)+1}$$. Our goal is to find $$g(x)$$, so we need to solve for it.

Solve for g(x)

Take the natural logarithm of both sides to simplify the expression and isolate g(x): $$\ln(x) = \ln\left(e^{3g(x)+1}\right)$$ Using the property of logarithms, we can move the exponent to the front: $$\ln(x) = (3g(x)+1)\ln(e)$$ Since $$\ln(e)=1$$, we have: $$\ln(x) = 3g(x)+1$$ Now, solve for $$g(x)$$: $$g(x) = \frac{\ln(x)-1}{3}$$

Differentiate g(x)

Now, we need to differentiate g(x) to find the derivative of the inverse function: $$\frac{d}{dx}\left(\frac{\ln(x)-1}{3}\right)$$ Using the quotient rule, we get: $$g'(x) = \frac{1}{3x}$$ Now, we have the derivative of the inverse function $$g'(x)=\frac{1}{3x}$$.

Key concepts

Derivative of Inverse Function

When we talk about the derivative of an inverse function, we are referring to the rate of change of one function that reverses the effect of another function. If a function \( f(x) \) is invertible, then its inverse function is denoted as \( f^{-1}(x) \). The derivative of this inverse function, \( f^{-1}'(x) \), can be used to determine how quickly or slowly the inverse function changes with respect to changes in its input.
To find the derivative of an inverse function, we use the formula:

• \( f^{-1}'(x) = \frac{1}{f'(f^{-1}(x))} \)

This formula signifies the reciprocal relationship between a function's derivative and its inverse's derivative, assuming each derivative exists. It's important to ensure that the function \( f \) is differentiable and its derivative is non-zero at points of interest.
In the example of \( f(x) = e^{3x+1} \), we found its inverse function by performing algebraic manipulations to express \( x \) in terms of \( g(x) \), where \( g(x) \) is the inverse function. This allowed us to differentiate \( g(x) \) and thus find \( g'(x) = \frac{1}{3x} \) as the derivative of the inverse function.

Logarithms

Logarithms are the inverse of exponential functions and are incredibly useful for solving equations involving exponents. The natural logarithm, denoted as \( \ln(x) \), specifically refers to the logarithm with base \( e \), where \( e \approx 2.71828 \), a constant often found in mathematics and natural sciences.
When solving for the inverse of an exponential function, such as \( f(x) = e^{3x+1} \), logarithms help us simplify the expression. By taking the natural logarithm of both sides of the equation \( x = e^{3g(x) + 1} \), we can tap into properties of logarithms to isolate our variable of interest, \( g(x) \).
The key properties of logarithms used here include:

• \( \ln(e^a) = a \) because \( \ln \) and exponential functions are inverses.
• \( \ln(ab) = \ln(a) + \ln(b) \) for separating logarithmic terms.
• \( \ln(a^b) = b\ln(a) \) for bringing exponents down as multipliers.

Thanks to these properties, solving \( \ln(x) = (3g(x) + 1) \) to find \( g(x) = \frac{\ln(x)-1}{3} \) becomes straightforward.

Exponential Functions

Exponential functions are a type of mathematical function in which a constant base is raised to a variable exponent. In general form, an exponential function is expressed as \( f(x) = a^{x} \), where \( a \) is a positive constant base and \( x \) is the exponent.
The particular exponential function in our example, \( f(x) = e^{3x+1} \), is based on the natural base \( e \). The function describes how quickly \( y \) values grow as \( x \) increases due to the compounding trait of \( e \).
Some core characteristics of exponential functions include:

• They always pass through the point \( (0,1) \) when not shifted or scaled.
• They are one-to-one, meaning each input has exactly one unique output, facilitating the finding of inverses.
• Their slopes become steeper as \( x \) increases because the rate of change increases.

In our exercise, understanding the exponential function helps us reason through its inverse. By acknowledging the monotonic increase of the function, we ensure its invertibility and correctly apply logarithmic operations to derive \( g(x) = \frac{\ln(x)-1}{3} \). This approach allows us to confidently determine the rate at which the inverse function changes with respect to \( x \).