Question

An invertible function \(f(x)\) is given along with a point that lies on its graph. Using Theorem 2.7.1, evaluate \(\left(f^{-1}\right)^{\prime}(x)\) at the indicated value. \(f(x)=\frac{1}{1+x^{2}}, x \geq 0\) Point \(=(1,1 / 2)\) Evaluate \(\left(f^{-1}\right)^{\prime}(1 / 2)\)

Short answer

\(-2\)

Step-by-step solution

Understand Theorem 2.7.1

Theorem 2.7.1 states that if a function \( f \) is invertible and differentiable at \( x \), and its inverse is differentiable at \( f(x) \), then the derivative of the inverse at \( f(x) \) is given by: \[ \left(f^{-1}\right)'(f(x)) = \frac{1}{f'(x)} \] We will apply this theorem to find \( \left(f^{-1}\right)'(\frac{1}{2}) \).

Determine the derivative of f(x)

The given function is \( f(x) = \frac{1}{1+x^{2}} \). Using the quotient rule, \[ f'(x) = \frac{(0)(1+x^2)-(1)(2x)}{(1+x^2)^2} = -\frac{2x}{(1+x^2)^2} \] This is the derivative of \( f(x) \).

Use the given point to find x

The point \((1, \frac{1}{2})\) on the graph of \( f \) indicates that \( f(1) = \frac{1}{2} \). Therefore, \( f^{-1}(\frac{1}{2}) = 1 \). This means that we look for \( \left(f^{-1}\right)'(\frac{1}{2}) \) using \( x = 1 \).

Evaluate the derivative of the inverse function

Using Step 1, we have: \[ \left(f^{-1}\right)'\left(\frac{1}{2}\right) = \frac{1}{f'(1)} \]From Step 2, \( f'(x) = -\frac{2x}{(1+x^2)^2} \). Substitute \( x = 1 \):\[ f'(1) = -\frac{2(1)}{(1+1^2)^2} = -\frac{2}{4} = -\frac{1}{2} \]Then, the derivative of the inverse is:\[ \left(f^{-1}\right)'\left(\frac{1}{2}\right) = \frac{1}{-\frac{1}{2}} = -2 \]

Key concepts

Derivative of Inverse Function

When tackling problems involving invertible functions, understanding the derivative of the inverse function is crucial. It's a way to find how rapidly the inverse function is changing, specifically at a given point.
A neat trick here is provided by Theorem 2.7.1. This theorem helps us calculate this derivative without directly finding the inverse function, which can sometimes be tricky or impossible to express cleanly.
Instead, if we have a function that is invertible and differentiable, and we know a specific point on its graph, we can easily use this theorem to find the required derivative value for its inverse at that particular point.

• Theorem 2.7.1 gives us: \[ \left(f^{-1}\right)'(f(x)) = \frac{1}{f'(x)} \] It tells us the relationship between derivatives of a function and its inverse.

If you understand this, finding the derivative of the inverse function becomes a more straightforward task.

Quotient Rule

The quotient rule is a fundamental tool in calculus used to differentiate functions expressed as a division of two differentiable functions. If you have a function in the form of \(g(x)/h(x)\), the derivative can be found using:

• \[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \]

In the context of the original exercise, \(f(x) = \frac{1}{1+x^{2}}\), the rule allows us to handle the differentiation even when dealing with more complex, multi-part functions. Here,

• Because the numerator was 1, its derivative was 0.
• The focus was on differentiating \(1+x^{2}\), which yielded \(2x\).

Thus, applying the quotient rule here simplifies the process, first yielding \(-\frac{2x}{(1+x^2)^2}\), which plays a key role in finding the derivative of the inverse function.

Theorem 2.7.1

Theorem 2.7.1 is an essential theorem in calculus, especially when dealing with invertible and differentiable functions. Often, students are asked to find the derivative of an inverse function at a particular point. To apply this theorem effectively:

• Ensure that the original function \(f(x)\) is both invertible and differentiable.
• Check that you have or can find the specific point where you want to evaluate the derivative of the inverse function.
• Compute the derivative of the original function \(f'(x)\).

By doing the above, you can apply the theorem's formula for the derivative of the inverse:\[ \left(f^{-1}\right)'(f(x)) = \frac{1}{f'(x)} \] This simplifies the process significantly. For the given point, if \(f(a) = b\), you can find \( (f^{-1})'(b) \) with \( a \). This theorem serves as a helpful shortcut around calculating inverse derivatives directly.

Differentiable Function

A differentiable function is simply a function that can be differentiated, which means one can calculate its derivative at a given point. In simpler terms, if a function is smooth and has no sharp corners or discontinuities, it is likely differentiable.
Differentiability plays a vital role when applying derivatives to real-world problems or complex calculus exercises. It's practically the key to making the calculations workable and reliable.

• In the problem from the exercise, the function \( f(x) = \frac{1}{1+x^{2}} \) is differentiable for all \(x\).
• This property allows us to use both the quotient rule to find \( f'(x) \), and then apply Theorem 2.7.1 reliably.

Before jumping into derivations, always check whether the function in question is differentiable within the domain you are analyzing. This ensures a correct and smooth calculation process, contributing to well-grounded mathematical solutions.