Problem 46 - Single Variable Calculus

Question

Let \(g\) denote the inverse of the function \(f\). (a) Show that the point \((a, b)\) lies on the graph of \(f .\) (b) Find \(g^{\prime}(b)\) $$ f(x)=\frac{1}{\sqrt{x^{2}+1}}, \text { where } x \geq 0 ; \quad\left(1, \frac{\sqrt{2}}{2}\right) $$

Step-by-Step Solution

Answer

The point (1, \(\frac{\sqrt{2}}{2}\)) lies on the graph of \(f(x)=\frac{1}{\sqrt{x^2+1}}\), and the derivative of its inverse function \(g(y)=\sqrt{\frac{1}{y^2} - 1}\) is \(g'(y)=\frac{-1}{y^3 \cdot 2 \sqrt{\frac{1}{y^2} - 1}}\). Therefore, \(g'\left(\frac{\sqrt{2}}{2}\right)=-2\).

Step 1: Verify the point (1, sqrt(2)/2) lies on the graph of f(x) #

We are given that f(x) = 1/√(x² + 1). To verify that the point (1, √2/2) lies on the graph of f, we need to substitute x = 1 into the function and see if we get the value √2/2. If f(1) is equal to \(\frac{\sqrt{2}}{2}\), then the point lies on the graph of f. $$ f(1) = \frac{1}{\sqrt{1^2 + 1}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$ Since f(1) = \(\frac{\sqrt{2}}{2}\), we can conclude that the point (1, \(\frac{\sqrt{2}}{2}\)) lies on the graph of f(x).

Step 2: Find the inverse function g(x) #

To find the inverse function g(x), we need to express x in terms of y from the given function f(x) = 1/√(x² + 1). Let y = f(x), so we have: \(y = \frac{1}{\sqrt{x^2+1}}\) Now we need to find x in terms of y. We can do this by using algebraic manipulation: \(\sqrt{x^2 + 1} = \frac{1}{y}\) \(x^2 + 1 = \frac{1}{y^2}\) \(x^2 = \frac{1}{y^2} - 1\) \(x= g(y) = \sqrt{\frac{1}{y^2} - 1}\) So the inverse function g(y) = \(\sqrt{\frac{1}{y^2} - 1}\).

Step 3: Differentiate the inverse function g'(y) #

Calculate the derivative of the inverse function g: $$ g'(y) = \frac{d}{dy} \left(\sqrt{\frac{1}{y^2} - 1}\right) $$ Let \(u = \frac{1}{y^2} - 1\), then g'(y) = \(\frac{d}{dy}(\sqrt{u})\). Using the chain rule, we have: $$ g'(y) = \frac{d\sqrt{u}}{dy} = \frac{\frac{d\sqrt{u}}{du} \cdot \frac{du}{dy}}{1} $$ Differentiating u with respect to y and using the power rule for derivatives: $$ \frac{du}{dy} = -2 \cdot \frac{1}{y^3} $$ Differentiating sqrt(u) with respect to u: $$ \frac{d\sqrt{u}}{du} = \frac{1}{2 \sqrt{u}} $$ Finally, applying the chain rule and replacing u with the original expression: $$ g'(y) = \frac{1}{2 \sqrt{\frac{1}{y^2} - 1}} \cdot -2 \cdot \frac{1}{y^3} $$ Simplifying the expression, we get: $$ g'(y) = \frac{-1}{y^3 \cdot 2 \sqrt{\frac{1}{y^2} - 1}} $$

Step 4: Find g'(sqrt(2)/2)

Finally, we substitute the value √2/2 into our derivative expression to find g'(√2/2): $$ g'\left(\frac{\sqrt{2}}{2}\right) = \frac{-1}{\left(\frac{\sqrt{2}}{2}\right)^3 \cdot 2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2}\right)^2} - 1}} $$ Simplify: $$ g'\left(\frac{\sqrt{2}}{2}\right) = \frac{-1}{\left(\frac{1}{2}\right)^3 \cdot 2 \sqrt{\frac{1}{\frac{1}{2}} - 1}} = \frac{-1}{\frac{1}{4} \cdot 2 \sqrt{1}} = \frac{-1}{\frac{1}{2}} = -2 $$ Thus, the value of the derivative of the inverse function g'(b) when b = √2/2 is -2.

Key Concepts

Derivative of Inverse Function

Understanding the derivative of an inverse function is crucial in calculus. When we have a function, say \(f(x)\), and its inverse denoted as \(g(y)\), the relationship between their derivatives can be expressed with the formula:

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

This essentially means that the derivative of the inverse function at a point \(b\) in its range is the reciprocal of the derivative of the original function \(f\) at the corresponding point \(a\) in its domain.
To find \(g'(b)\), you need the derivative of \(f\) evaluated at the point \(x\) such that \(f(x) = b\). In the step-by-step solution above, after verifying and finding the point on \(f(x)\), we determine \(g'(\frac{\sqrt{2}}{2}) = -2\). This process involves both evaluating \(f'(x)\) and using it to find \((f^{-1})'(b)\). Remember, knowing these relationships allows you to efficiently compute derivatives when inverse functions are involved.

Chain Rule

The chain rule is a staple in differentiation used when dealing with composite functions. It states that if you have two functions, \(u(y)\) being a function of \(y\), and \(g(u)\) being a function of \(u\), then the derivative of \(g(u(y))\) with respect to \(y\) is:

• \(\frac{d}{dy}[g(u(y))] = g'(u) \cdot u'(y)\)

In the context of the step-by-step solution, we used the chain rule to differentiate the inverse function \(g(y) = \sqrt{\frac{1}{y^2} - 1}\).
This involved two main steps:

• First, define an inner function, \(u = \frac{1}{y^2} - 1\), and then differentiate with respect to \(y\).
• Next, differentiate the outer function, \(\sqrt{u}\), with respect to \(u\), resulting in \(\frac{1}{2\sqrt{u}}\).

The final derivative, \(g'(y)\), was obtained by multiplying these derivatives according to the chain rule, making it a vital tool for solving the given problem.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying expressions to make them easier to work with. In this exercise, we utilized algebraic manipulation to express \(x\) as a function of \(y\), invert \(f(x)\), and ultimately find the inverse function \(g(y)\).First, we started with the equation \(y = \frac{1}{\sqrt{x^2+1}}\).

• By setting \(y = \frac{1}{\sqrt{x^2+1}}\), we isolated \(\sqrt{x^2 + 1}\) on one side: \(\sqrt{x^2 + 1} = \frac{1}{y}\).
• This was squared to eliminate the square root, giving \(x^2 + 1 = \frac{1}{y^2}\).
• Next, solving for \(x^2\) gave \(x^2 = \frac{1}{y^2} - 1\), and then \(x = \sqrt{\frac{1}{y^2} - 1}\).

These steps show how algebraic manipulation allows us to derive \(g(y)\), the inverse of \(f(x)\), which is pivotal for finding the derivative of the inverse function as performed in the solution. Mastering these techniques is essential for solving complex problems in calculus.