Problem 6 - Calculus Early Transcendentals

Question

Explain how to find \(\left(f^{-1}\right)^{\prime}\left(y_{0}\right),\) given that \(y_{0}=f\left(x_{0}\right)\)

Step-by-Step Solution

Answer

Question: Given that \(y_0 = f(x_0)\), find the derivative of the inverse function \((f^{-1})'(y_0)\) using the inverse function theorem. Answer: To find the derivative of the inverse function \((f^{-1})'(y_0)\), follow these steps: 1) Determine the derivative of the original function, \(f'(x)\). 2) Use the inverse function theorem to find \((f^{-1})'(y_0)\) by taking the reciprocal of the derivative of the original function evaluated at \(x_0\): \((f^{-1})'(y_0) = \frac{1}{f'(x_0)}\).

Step 1: Find the derivative of the original function, \(f'(x)\)

To find the derivative of the original function, we need to differentiate it with respect to \(x\). Given that \(y_0 = f(x_0)\), we differentiate both sides of this equation with respect to \(x\) to find \(f'(x)\): $$\frac{dy}{dx} = f'(x)$$

Step 2: Use the inverse function theorem to find the derivative of the inverse function, \((f^{-1})'(y_0)\)

According to the inverse function theorem, the derivative of the inverse function \((f^{-1})'(y_0)\) can be found by taking the reciprocal of the derivative of the original function, evaluated at \(x_0\): $$(f^{-1})'(y_0) = \frac{1}{f'(x_0)}$$ We know that \(y_0 = f(x_0)\), so we can find the value of \(f'(x_0)\) by substituting \(x_0\) into the expression for \(f'(x)\) that we found in Step 1. Once we have the value of \(f'(x_0)\), we can find the derivative of the inverse function by taking the reciprocal of this value: $$(f^{-1})'(y_0) = \frac{1}{f'(x_0)}$$ By following these two steps, we have found the derivative of the inverse function \((f^{-1})'(y_0)\).

Key Concepts

Derivative

A derivative is a crucial concept in calculus that describes how a function changes as its input changes. It is essentially the rate at which a function is changing at any given point. To compute the derivative of a function, you apply differentiation, which involves finding the limit of the average rate of change of the function over a very small interval. For example, the derivative of a function, denoted as \( f'(x) \), tells us the slope of the tangent line to the curve of the function at any point \( x \). This can be particularly useful for understanding how the function behaves locally and predicting its behavior in a small region.When you have a specific point, like \( x_0 \), you evaluate the derivative at that point to find \( f'(x_0) \). This gives the slope of the function at \( x_0 \), which is essential when applying the inverse function theorem. The value of \( f'(x_0) \) is used in calculating the derivative of the inverse function, showing how intertwined these mathematical concepts are.

Inverse Function

An inverse function reverses the operation of the original function. If you have a function \( f(x) \), its inverse \( f^{-1}(x) \) will give you back the original input when applied to the output of \( f(x) \). In simpler terms, if \( y = f(x) \), then \( x = f^{-1}(y) \). Understanding the inverse function is pivotal in solving equations where the function has been applied, as it enables you to backtrack to the original value.An important aspect of inverse functions is understanding when they exist. For a function to have an inverse, it usually needs to be one-to-one (bijective) so that each output is assigned to exactly one input. This ensures the function's graph passes the horizontal line test, which visually confirms that no horizontal line intersects the curve more than once. Once an inverse is found, its properties allow us to explore the rates of change through the derivative of this inverse, which we determine using the inverse function theorem.

Reciprocal

A reciprocal is something you might remember from fraction arithmetic: it's simply one over a number. In calculus, the reciprocal concept is extended to derivatives when considering inverse functions. The inverse function theorem states that if a function \( f \) is invertible around a point and differentiable, all derivatives involved are continuous, then the derivative of its inverse function \( (f^{-1})'(y_0) \) is the reciprocal of the derivative of the function at the corresponding point. Mathematically, this concept can be expressed as:- \((f^{-1})'(y_0) = \frac{1}{f'(x_0)}\)This equation is important because it provides a direct method to compute the derivative of an inverse function using the derivative of the original function. The simplicity of taking a reciprocal helps to avoid more complex differentiation procedures, making the process both efficient and elegant.