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Chapter 3: Problem 15

Use the Inverse Function Derivative Rule to calculate \(\left(f^{-1}\right)^{\prime}(t)\). $$ f:(1,3) \rightarrow(3,15), f(s)=s^{2}+2 s $$

Short Answer

Expert verified
The derivative \( \left(f^{-1}\right)^{\prime}(t) \) is \( \frac{1}{2\sqrt{1+t}} \).

Step by step solution

01

Understand the Inverse Function Derivative Rule

Before we dive into calculations, let's understand what the Inverse Function Derivative Rule is. If a function \( f(x) \) has an inverse \( f^{-1}(x) \) and both are differentiable, then the derivative of the inverse function at a point \( t \) is given by the formula: \( \left(f^{-1}\right)^{\prime}(t) = \frac{1}{f^{\prime}(s)} \), where \( s = f^{-1}(t) \). The idea is to find \( s \) first such that \( f(s) = t \).
02

Express \( f(s) \) and Identify \( t \)

The function is given by \( f(s) = s^2 + 2s \) and it maps from \( (1,3) \) to \( (3,15) \). We are tasked to find the derivative \( (f^{-1})'(t) \). Here \( t \) is a specific value within the range (3,15). Let's assume \( t = f(s) \) for simplifying calculations.
03

Solve for \( s \) such that \( f(s) = t \)

To find \( s \) we need \( f(s) = s^2 + 2s = t \). This is a quadratic equation in the form \( s^2 + 2s - t = 0 \). Solving this equation for \( s \) will give us: \( s = \frac{-2 \pm \sqrt{4 + 4t}}{2} = -1 \pm \sqrt{1+t} \). The value of \( s \) must be in the interval (1,3), so we choose \( s = -1 + \sqrt{1+t} \).
04

Find the Derivative \( f^{\prime}(s) \)

We need to find the derivative of \( f(s) \) to use it in the Inverse Function Derivative Rule. The derivative \( f^{\prime}(s) \) is: \( f^{\prime}(s) = \frac{d}{ds}(s^2 + 2s) = 2s + 2 \).
05

Apply the Inverse Function Derivative Formula

Now we substitute in the values to find \( \left(f^{-1}\right)^{\prime}(t) \). Using the derivative formula, \( \left(f^{-1}\right)^{\prime}(t) = \frac{1}{2s + 2} \). Substitute the expression for \( s = -1 + \sqrt{1+t} \), we get: \( \left(f^{-1}\right)^{\prime}(t) = \frac{1}{2(-1 + \sqrt{1+t}) + 2} \). Simplify to find the final expression.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a fundamental concept in calculus that allows us to understand how functions change. In the context of the exercise, differentiation is used to find the rate of change of a function. Here is a brief overview:
  • Derivatives tell us how a function behaves as its input changes. Essentially, they give us the slope of the function at any given point.
  • For the function \( f(s) = s^2 + 2s \), the derivative \( f'(s) \) measures the rate of change or the slope of the function with respect to \( s \).
  • To find \( f'(s) \), apply basic differentiation rules: for any terms \( as^n \), the derivative is \( na s^{n-1} \).
In our example, the derivative of \( f(s) \), calculated as \( 2s + 2 \), is crucial for finding the derivative of the inverse function. Differentiation provides the insight into how \( f(s) \) behaves, which is necessary for understanding its inverse.
Quadratic Equations
A quadratic equation takes the form \( ax^2 + bx + c = 0 \), and solving it is essential for tasks such as finding specific points on a curve. In this problem, the original function \( f(s) = s^2 + 2s \) leads to the quadratic equation \( s^2 + 2s - t = 0 \). Quadratic equations can be solved using several methods:
  • Factoring: Works if the equation can be expressed as a product of two binomials.
  • Completing the square: Rewriting the equation to isolate \( s^2 \).
  • Quadratic formula: Always applicable, given by \( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
In our pivotal step, solving the quadratic equation gave us \( s = -1 \pm \sqrt{1+t} \). Given the constraint that \( s \) must be within the interval (1, 3), we select \( s = -1 + \sqrt{1+t} \). This value of \( s \) is crucial for applying the inverse function derivative rule.
Derivatives of Inverse Functions
The inverse function derivative rule is an important tool in calculus, allowing us to find derivatives of inverse functions without directly differentiating the inverse itself. This rule states:\[\left(f^{-1}\right)'(t) = \frac{1}{f'(s)}\]Where \( s = f^{-1}(t) \), meaning \( f(s) = t \).Key points to remember:
  • Existence: The function \( f(x) \) must have an inverse, meaning it has to be one-to-one.
  • Differentiability: Both \( f(x) \) and its inverse must be differentiable.
  • Calculation: You find \( s \) such that \( f(s) = t \), then calculate the derivative \( f'(s) \).
  • Application: The formula \( \left(f^{-1}\right)'(t) = \frac{1}{2(-1 + \sqrt{1+t}) + 2} \) offers a simplified way to find the rate of change for the inverse function.
In our example, we've identified and simplified the expression for \( \left(f^{-1}\right)'(t) \) once the value of \( s \) was known, providing insights into the behavior of the inverse function relative to \( t \).

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Most popular questions from this chapter

\- A real-valued function \(f\) of a real variable \(x\) is said to be algebraic if there is a polynomial \(p(u, v)\) with integer coefficients such that \(p(x, f(x))=0\) for all \(x .\) For example, \(f(x)=\) \(2 x+\sqrt{x^{2}+1}\) is algebraic because $$ \begin{aligned} p(x, f(x))=& 3 x^{2}+\left(2 x+\sqrt{x^{2}+1}\right)^{2} \\ &-4 x\left(2 x+\sqrt{x^{2}+1}\right)-1 \\ \equiv & 0 \end{aligned} $$ for \(p(u, v)=3 u^{2}+v^{2}-4 u v-1 .\) A function that is not algebraic is said to be transcendental. Find a polynomial that shows that the given expression is algebraic. $$ \sqrt[3]{x+\sqrt{x}} $$

Verify the identity $$ \tanh (x+y)=\frac{\tanh (x)+\tanh (y)}{1+\tanh (x) \tanh (y)} $$

A function \(f,\) a point \(c,\) an increment \(\Delta x,\) and a positive integer \(n\) are given. Use the method of increments to estimate \(f(c+\Delta x)\). Then let \(h=\Delta x\) / \(N\). Use the method of increments to obtain an estimate \(y_{1}\) of \(f(c+h) .\) Now, with \(c+h\) as the base point and \(y_{1}\) as the value of \(f(c+h),\) use the method of increments to obtain an estimate \(y_{2}\) of \(f(c+2 h)\). Continue this process until you obtain an estimate \(y_{N}\) of \(f(c+N \cdot h)=f(c+\Delta x) .\) We say that we have taken \(N\) steps to obtain the approximation. The number \(h\) is said to be the step size. Use a calculator or computer to evaluate \(f(c+\Delta x)\) directly. Compare the accuracy of the one step and \(N\) -step approximations. $$ f(x)=\sqrt{x}, c=4, \Delta x=0.5, N=5 $$

Show that \(y=\sinh (\omega x)\) and \(y=\cosh (\omega x)\) are solutions of the differential equation \(y^{\prime \prime}(x)-\omega^{2} y(x)=0\). Let \(a\) be a positive constant, and set $$ T(x)=a \cdot \sec h^{-1}\left(\frac{x}{a}\right)-\sqrt{a^{2}-x^{2}} $$

Calculate the value of the given inverse trigonometric function at the given point. $$ \arccos (\sqrt{2} / 2) $$
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