Question

Consider the following functions (on the given interval, if specified). Find the derivative of the inverse function. $$f(x)=x^{2}-4, \text { for } x>0$$

Short answer

Answer: The derivative of the inverse function is \((f^{-1}(y))' = \frac{1}{2\sqrt{y+4}}\).

Step-by-step solution

Find the inverse function

To find the inverse function, we must first solve for \(x\) in the equation \(y = x^2 - 4\). $$y = x^2 - 4$$ $$x^2 = y + 4$$ $$x = \pm\sqrt{y + 4}$$ However, since the initial condition states that \(x>0\), we will only consider the positive square root. Therefore, the inverse function, \(f^{-1}(y)\), is: $$f^{-1}(y)= \sqrt{y+4}$$

Find the derivative of the inverse function

Now we need to find the derivative of the inverse function. The formula for the derivative of an inverse function is: $$(f^{-1}(y))' = \frac{1}{f'(f^{-1}(y))}$$ First, we need to find the derivative of the given function \(f(x)=x^2-4\): $$f'(x) = 2x$$ Next, we substitute the inverse function into the derivative of the given function: $$f'(f^{-1}(y)) = 2f^{-1}(y) = 2\sqrt{y+4}$$ Now we can find the derivative of the inverse function: $$(f^{-1}(y))' = \frac{1}{f'(f^{-1}(y))} = \frac{1}{2\sqrt{y+4}}$$ So the derivative of the inverse function is: $$(f^{-1}(y))' = \frac{1}{2\sqrt{y+4}}$$

Key concepts

Inverse Function

Understanding inverse functions is a fundamental concept in mathematics, particularly within calculus. An inverse function, denoted as \(f^{-1}(x)\), effectively reverses the action of the original function \(f(x)\). In other words, it provides an output that, when fed back into the original function, will return the input variable of the inverse function.

For instance, if we have a function that squares a number, the inverse function would extract the square root of that number. However, it's important to ensure that the function is one-to-one, which means that for every output there's a unique input. This is critical because without this property, a function would not have a well-defined inverse.

When attempting to derive the inverse of a function, one commonly interchanges the roles of \(x\) and \(y\), and then solves for \(x\). This process can be tricky, especially in regards to domain restrictions which must be considered to ensure the function remains one-to-one. For example, in the exercise where the function \(f(x) = x^2 - 4\) has a restriction \(x > 0\), the positive square root is the correct branch of the inverse to avoid a multi-valued function.

Calculus

Calculus is a significant branch of mathematics that deals with the properties and applications of derivatives and integrals. It's used to understand changes between values that are related by a function. There are two primary divisions within calculus: differential calculus, which concerns the concept of a derivative, and integral calculus, which focuses on the concept of an integral.

The fundamental concept of calculus is the limit, the value that a sequence or function approaches as the index or input approaches some value. Limits help define the precise calculation of the slope of a line tangent to a curve (the derivative) and the area under a curve (the integral). Calculus is not only critical for mathematics majors but also essential in fields like physics, engineering, economics, statistics, and more, as it provides the tools to model and analyze dynamic systems.

Derivative Calculation

The derivative is a principal concept in calculus representing the rate at which a quantity changes. It's the foundational tool to understand motion, growth, and decay and is generally a measure of how a function changes as its input changes.

Mathematically, the derivative of a function \(f\) at a particular point \(x\) is the limit of the average rate of change of \(f\) in a small interval around \(x\). The notation \(f'(x)\) or \(\frac{dy}{dx}\) is often used to denote the derivative of \(f\) with respect to \(x\). To compute a derivative, we apply specific rules, such as the power rule, product rule, quotient rule, and the chain rule, depending on the composition of the function.

When finding the derivative of an inverse function, as in our textbook exercise, we use the reciprocal of the derivative of the original function at the corresponding point. This requires first finding the inverse function, then the derivative of the original function, and finally applying the formula \(\frac{1}{f'(f^{-1}(y))}\). It's crucial to apply these steps methodically to ensure accuracy in the derivative calculation for inverse functions.