Problem 10 - Calculus - AP Edition

Question

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

Step-by-Step Solution

Answer

The derivative of the inverse function is \( \frac{1}{5 - 2\cos 2x} \).

Step 1: Understand the Relationship between a Function and its Inverse

The inverse function theorem states that if a function \( f \) is differentiable at \( x \) and its derivative \( f'(x) \) is non-zero, then its inverse \( f^{-1} \) is also differentiable at \( f(x) \), and \ \[ \left(f^{-1}\right)'(y) = \frac{1}{f'(x)} \] where \( y = f(x) \).

Step 2: Differentiate the Given Function

Differentiate the function \( f(x) = 5x - \sin 2x \). The derivative \( f'(x) \) is obtained as follows: \ \[ f'(x) = 5 - 2\cos 2x \] using the chain rule for differentiating the \( \sin 2x \) component.

Step 3: Apply the Inverse Function Derivative Formula

Using the inverse function theorem, the derivative of the inverse function \( \left(f^{-1}\right)'(y) \) at \( y \) where \( y = f(x) \) is \ \[ \left(f^{-1}\right)'(y) = \frac{1}{5 - 2\cos 2x} \].

Step 4: Implicit Differentiation Method

Starting from the equation \( y = 5x - \sin 2x \), differentiate both sides with respect to \( y \). \ 1. Differentiate implicitly to get: \ \( \frac{dy}{dy} = 5\frac{dx}{dy} - 2\cos 2x \frac{dx}{dy} \). \ 2. Set \( 1 = \left(5 - 2\cos 2x\right) \frac{dx}{dy} \) and solve for \( \frac{dx}{dy} \): \ \( \frac{dx}{dy} = \frac{1}{5 - 2\cos 2x} \).

Step 5: Verify and Conclusion

The result from the implicit differentiation in Step 4 matches the result from the inverse theorem in Step 3: \( \left(f^{-1}\right)'(y) = \frac{1}{5 - 2\cos 2x} \). This confirms that the derivative of the inverse function is correctly calculated.

Key Concepts

Derivative

A derivative represents how a function changes as its input changes. In simpler terms, it tells us the rate at which a quantity is changing. When calculating the derivative, we seek the slope of the function at any point. For instance, the derivative of a linear function like

• \( f(x) = 5x - \sin 2x \)

relates to how this function changes as \( x \) changes.
The derivative itself, \( f'(x) \), gives us

• \( 5 - 2\cos 2x \)

This derivative tells us the function's rate of change is a combination of the constant rate from 5 and a variable rate from the \( \cos \) function in the \( \sin \) term. Understanding derivatives is essential, not just for solving equations, but for modeling real-world scenarios where change is involved. For example, speed can be thought of as the derivative of distance with respect to time.

Implicit Differentiation

Implicit differentiation involves differentiating functions that are not given explicitly, meaning they are not isolated as \( y = f(x) \). In our case, we have an equation describing \( y \) as a function of \( x \),

• \( y = 5x - \sin 2x \)

To find \( \frac{dx}{dy} \), the rate at which \( x \) changes with \( y \):

• We differentiate each term with respect to \( y \), keeping in mind how each variable relates.
• Why implicit? Sometimes, isolating \( y \) isn't feasible, needing implicit differentiation to work through it.

This method allowed us to find \( \frac{dx}{dy} = \frac{1}{5 - 2\cos 2x} \). Using implicit differentiation is powerful when you can't or don't need to express a function in straightforward \( y = ... \) form, yet still need derivatives.

Chain Rule

The chain rule is a fundamental tool for finding derivatives of composite functions. It deals with functions within functions, known as nested functions. For instance, in our function:

• \( f(x) = 5x - \sin 2x \)

The \( \sin \) term is a function inside the larger function. Here, the chain rule helps in dealing with the derivative of \( \sin 2x \).

• How it works: First, differentiate the outer function while keeping the inner one intact. Then, multiply by the derivative of the inner function. For \( \sin 2x \), it becomes \( 2\cos 2x \).
• Why use it? Without it, calculating derivatives of more elaborate functions wouldn't be possible.

By learning the chain rule, you gain a tool to tackle functions embedded within others, enhancing your ability to deal with complex mathematical expressions.