Problem 77 - Calculus Single Variable

Question

Suppose that \(f\) is invertible and twice differentiable, that \(f(c)=\gamma,\) and that \(f^{\prime}(c) \neq 0 .\) Show that $$\left(f^{-1}\right)^{\prime \prime}(\gamma)=-\frac{f^{\prime \prime}(c)}{\left(f^{\prime}(c)\right)^{3}}$$

Step-by-Step Solution

Answer

The second derivative of the inverse function is \( \left(f^{-1}\right)^{\prime\prime}(\gamma) = -\frac{f^\prime\prime(c)}{(f^\prime(c))^3} \).

Step 1: Express the Inverse Derivative

Given that \( f(c) = \gamma \), we know that \( f^{-1}(\gamma) = c \). Using the formula for the derivative of an inverse function, we have \( \left(f^{-1}\right)'(\gamma) = \frac{1}{f'(c)} \). This is the first derivative of the inverse function.

Step 2: Differentiate the Inverse Function Again

To find the second derivative, \( \left(f^{-1}\right)''(\gamma) \), we differentiate \( \left( f^{-1} \right)'(\gamma) = \frac{1}{f'(c)} \) with respect to \( \gamma \). This requires using the chain rule since \( c = f^{-1}(\gamma) \).

Step 3: Apply the Chain Rule

Using the chain rule, differentiate \( \left( f^{-1} \right)'(\gamma) = \frac{1}{f'(c)} \) with respect to \( \gamma \) gives \( \left( f^{-1} \right)''(\gamma) = \frac{d}{d\gamma} \left( \frac{1}{f'(c)} \right) = -\frac{f''(c)}{(f'(c))^3} \). This applies the derivative of \( \frac{1}{u} \) being \( -\frac{u'}{u^2} \) as part of the application of the chain rule.

Step 4: Final Expression and Conclusion

Thus, the expression for the second derivative is \( \left( f^{-1} \right)''(\gamma) = - \frac{f''(c)}{(f'(c))^3} \) which confirms the given condition.

Key Concepts

Inverse Function

An inverse function essentially reverses the operations of the original function. If you start with the function \( f(x) \) and you get a value \( y \), the inverse function \( f^{-1}(y) \) will take you back to \( x \). For \( f \) to have an inverse, it needs to be bijective; this means it has to be both one-to-one and onto.

• "One-to-one" ensures that each output is determined by only one input.
• "Onto" means that every possible output is covered by the function.

So, when we say \( f(c) = \gamma \), the inverse function should give us \( f^{-1}(\gamma) = c \). This property helps us map outputs back to inputs in consistent and predictable ways. Understanding inverse functions is crucial for solving equations and modeling real-world situations.

Chain Rule

The chain rule is a technique in calculus used for finding the derivative of a composite function. A composite function is when you have one function nested inside another, like \( h(x) = g(f(x)) \). To differentiate \( h(x) \), the chain rule states you should:

• Differentiate the outer function \( g \, \), leaving the inner function \( f(x) \) unchanged.
• Multiply by the derivative of the inner function \( f(x) \).

Formally, this can be written as \( h'(x) = g'(f(x)) \cdot f'(x) \). By using the chain rule cleverly, we find derivatives of complex functions by breaking them down into simpler parts. The chain rule is fundamental in advanced derivative calculations, like finding the inverse derivative or second derivative.

Derivative

A derivative measures how a function changes as its input changes, encapsulating the idea of the rate of change. If you consider the function \( f(x) \), the derivative, \( f'(x) \), can be interpreted as the slope of the tangent line to the curve at any point \( x \).

• It tells us how steep the curve is at a specific point.
• Mathematically, it's the limit of \( \frac{f(x+h) - f(x)}{h} \) as \( h \to 0 \).

The derivative is a crucial concept as it helps in understanding alterations within a function, like acceleration for velocity functions. Knowledge of derivatives is the main tool for analyzing functions in almost every calculus topic.

Inverse Derivative

To find the derivative of an inverse function \( f^{-1} \), we use a particular formula involving the original function. If \( f \) is differentiable and its inverse exists, then the derivative of the inverse is given by:

• \( \left(f^{-1}\right)'(\gamma) = \frac{1}{f'(c)} \)

This formula arises from rearranging terms related to the derivative definition in the context of inverse functions. It signifies that if the derivative of \( f \) is large, the derivative of its inverse will be small, and vice versa. The calculations of an inverse derivative are profound in fields such as physics, where understanding the inversion of a function's behavior is essential. It's a specialized tool, crucial in scenarios where the inverse operations must be precisely quantified.

Second Derivative

The second derivative is the derivative of the derivative, represented as \( f''(x) \). It offers insight into the acceleration of the function or how the rate of change itself is changing.

• If \( f'(x) \) is about slopes, \( f''(x) \) delves into the curvature of the curve.
• When \( f''(x) > 0 \), the function is concave up (like a bowl), and when \( f''(x) < 0 \), the function is concave down (like a hill).

For inverse functions, finding the second derivative, such as \( \left(f^{-1}\right)''(\gamma) \), often involves using formulas reliant on both the original and inverse functions. These calculations are key to understanding the nature and shape of function graphs beyond simple rates of change, offering detailed insights into dynamic behaviors of mathematics in real-world applications.