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Chapter 3: Problem 10

Find the derivative of \(f^{-1}\) by using Formula (3), and check your result by differentiating implicitly. $$ f(x)=5 x-\sin 2 x, \quad-\frac{\pi}{4}

Short Answer

Expert verified
The derivative of \( f^{-1} \) is \( \frac{1}{5 - 2\cos(2x)} \). Implicit differentiation confirms this result.

Step by step solution

01

Differentiate Function f

The original function is given by \( f(x) = 5x - \sin(2x) \). The first step is to find the derivative of \( f \), which we will call \( f'(x) \). Differentiating term by term, we have:1. The derivative of \( 5x \) is 5.2. The derivative of \(-\sin(2x)\) is \(-2\cos(2x)\), using the chain rule.Therefore, the derivative of \( f \) is:\[ f'(x) = 5 - 2\cos(2x) \]
02

Use the Inverse Function Theorem

According to the inverse function theorem, if \( y = f(x) \) and \( f \) is differentiable and has an inverse, then the derivative of \( f^{-1}(y) \) is given by:\[ \left( f^{-1} \right)'(y) = \frac{1}{f'(x)} \]where \( y = f(x) \). Substituting the derivative \( f'(x) = 5 - 2\cos(2x) \) from Step 1, we have:\[ \left( f^{-1} \right)'(y) = \frac{1}{5 - 2\cos(2x)} \]
03

Choose a Point to Verify

Choose a specific point \( x_0 \) to verify the inverse derivative formula. Let's consider \( x_0 = 0 \). For \( x = 0 \), the function becomes \( f(0) = 5 \cdot 0 - \sin(0) = 0 \). Thus, \( y = 0 \).
04

Check the Derivative Using Implicit Differentiation

Using implicit differentiation on \( y = 5x - \sin(2x) \), differentiate both sides with respect to \( y \). This gives:1. \( dy/dy = 1 \)2. \( d/dy(5x) = 5 \cdot dx/dy \)3. \( d/dy(-\sin(2x)) = -2\cos(2x)\cdot dx/dy \)Combining these, we have:\[ 1 = 5 \cdot \frac{dx}{dy} - 2\cos(2x) \cdot \frac{dx}{dy} \]Simplifying, \( \frac{dx}{dy} = \frac{1}{5 - 2\cos(2x)} \).This verifies our result from Step 2.

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Key Concepts

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

Derivative
A derivative represents the rate at which a function changes as its input changes. In simpler terms, it measures how the function f's y-value changes as the x-value changes. Let's take the function in our example, \( f(x) = 5x - \sin(2x) \).

To find the derivative \( f'(x) \), we differentiate each term:
  • The derivative of \( 5x \) is 5. This is because the derivative of \( ax \) is just \( a \), where \( a \) is a constant.
  • The derivative of \(-\sin(2x)\) requires the chain rule. The derivative of \(-\sin(u)\) is \(-\cos(u) \cdot u'\), where \( u = 2x \). So, differentiate \( 2x \) to get 2, and hence, the derivative becomes \(-2\cos(2x)\).
Thus, the derivative of the function is \( f'(x) = 5 - 2\cos(2x) \). Derivatives are foundational in calculus as they help to understand how a function behaves and changes. They can describe motion, growth, and many other real-life phenomena.
Inverse Function Theorem
The inverse function theorem helps us find the derivative of the inverse of a function. Basically, if a function \( f \) is differentiable and its inverse \( f^{-1} \) exists near some point, the theorem gives us a formula to find \( (f^{-1})'(y) \).

According to the theorem:
  • If \( y = f(x) \), then \( (f^{-1})'(y) = \frac{1}{f'(x)} \).
  • This means, to find the derivative of \( f^{-1}(y) \), you take the reciprocal of \( f'(x) \).
In our example, the function \( f(x) = 5x - \sin(2x) \) has a derivative \( f'(x) = 5 - 2\cos(2x) \). Thus, the derivative of its inverse is \( (f^{-1})'(y) = \frac{1}{5 - 2\cos(2x)} \). This is a powerful tool because it lets us find derivatives of complex inverse functions easily.
Implicit Differentiation
Implicit differentiation is a technique used when dealing with equations not easily solved for one variable in terms of the other. Sometimes, equations relate variables in a way that isn鈥檛 straightforward, like \( y = 5x - \sin(2x) \).

To differentiate implicitly:
  • Differentiate each term of the equation with respect to \( y \), assuming \( x \) is a function of \( y \).
  • Apply the chain rule wherever necessary.
In the problem, differentiating \( 5x - \sin(2x) = y \) with respect to \( y \) gives:
  • \( 1 = 5 \cdot \frac{dx}{dy} - 2\cos(2x) \cdot \frac{dx}{dy} \).
Solving for \( \frac{dx}{dy} \) confirms \( \frac{dx}{dy} = \frac{1}{5 - 2\cos(2x)} \), exactly what we found using the inverse function theorem. Implicit differentiation is essential when tackling more involved equations in calculus.

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Most popular questions from this chapter

Determine whether the statement is true or false. Explain your answer. If \(y=f(x)\) is a function such that \(d y / d x\) is a rational function, then \(f(x)\) is also a rational function.

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Find the differential \(d y\). (a) \(y=x \sqrt{1-x}\) (b) \(y=(1+x)^{-17}\)
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