Problem 9 - Calculus Early Transcendentals

Question

Evaluate the derivatives of the following functions. $$f(w)=\cos \left(\sin ^{-1} 2 w\right)$$

Step-by-Step Solution

Answer

Answer: The derivative of the function with respect to \(w\) is \(\frac{df(w)}{dw} = \frac{-4w}{\sqrt{1-(2w)^2}}\).

Step 1: Identify the outer and inner functions

In this case, the outer function is the cosine function \(f(u) = \cos(u)\) and the inner function is the inverse sine function multiplied by 2, \(g(w) = \sin^{-1}(2w)\).

Step 2: Find the derivatives of the outer and inner functions

Find the derivatives of the outer and inner functions with respect to their respective variables: For the outer function, find the derivative of \(f(u)\) with respect to \(u\): $$f'(u) = -\sin(u)$$ For the inner function, find the derivative of \(g(w)\) with respect to \(w\): $$g'(w) = \frac{d}{dw}\left(\sin^{-1}(2w)\right) = \frac{2}{\sqrt{1 - (2w)^2}}$$ This is because the derivative of \(\sin^{-1}(x)\) is \(\frac{1}{\sqrt{1 - x^2}}\) and we also need to apply the chain rule to account for the multiplication by 2.

Step 3: Apply the chain rule

Now apply the chain rule to find the derivative of the given function with respect to \(w\): $$\frac{df(w)}{dw} = (f' \circ g) \cdot g'(w),$$ Substitute the previously found expressions for \(f'(u)\) and \(g'(w)\): $$\frac{df(w)}{dw} = \left(-\sin\left(\sin^{-1}(2w)\right)\right) \cdot \left(\frac{2}{\sqrt{1 - (2w)^2}}\right)$$

Step 4: Simplify the final expression

Simplify the expression: $$\frac{df(w)}{dw} = -\left(2w\right)\left(\frac{2}{\sqrt{1 - (2w)^2}}\right) = \frac{-4w}{\sqrt{1-(2w)^2}}$$ We have found the derivative of the given function with respect to \(w\): $$\frac{df(w)}{dw} = \frac{-4w}{\sqrt{1-(2w)^2}}$$

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus that allows us to find the derivative of a composite function. A composite function is created when one function is nested inside another, such as in the example given where we have a cosine function wrapping around an inverse sine function.
To apply the chain rule, we need to:

• Identify the outer and inner functions.
• Find the derivative of the outer function with respect to its variable (often denoted as a substitution variable).
• Find the derivative of the inner function with respect to its variable.
• Multiply these derivatives together according to the chain rule formula: \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\).

In this exercise, we see the outer function as \(f(u) = \cos(u)\) and the inner function as \(g(w) = \sin^{-1}(2w)\). The derivatives are critical because they connect the changes occurring at different layers of the composite function. In our example, the derivatives are:

• Outer function: \(f'(u) = -\sin(u)\)
• Inner function: \(g'(w) = \frac{2}{\sqrt{1 - (2w)^2}}\)

These derivatives are combined by the chain rule to give the derivative of the entire function. Understanding how each piece fits together is the backbone of mastering calculus and the chain rule is often the first step in unraveling the complexities of composite functions.

Inverse Trigonometric Functions

Inverse trigonometric functions provide angles corresponding to trigonometric ratios. Among them, \(\sin^{-1}(x)\) (also known as arcsine) gives the angle whose sine is \(x\). In calculus, finding the derivative of inverse trigonometric functions is crucial for solving problems involving rates of change.For \(\sin^{-1}(x)\), the derivative is given by:\[\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}}\]This formula reflects the rate at which the angle changes as the sine value changes.
In our problem, we take it a step further. The function inside the arcsine is scaled by 2, i.e., \(\sin^{-1}(2w)\). To find its derivative, we employ the chain rule. We adjust the standard derivative to account for this scaling: \[\frac{d}{dw}[\sin^{-1}(2w)] = \frac{2}{\sqrt{1-(2w)^2}}\]The factor of \(2\) comes from the derivative of the linear term \(2w\) with respect to \(w\). This adjustment is what makes calculus powerful in dealing with more complex expressions and nested functions. Understanding these derivatives is vital as they frequently appear in physics and engineering, where angles and rotations are key players.

Composite Functions

Composite functions are combinations of two or more functions wherein one function's output becomes the input for another. In the given example, \(f(w) = \cos(\sin^{-1}(2w))\), the cosine function takes as input the result of the inverse sine function. The beauty of composite functions lies in their ability to represent more complex relationships between variables.When dealing with derivatives of composite functions, like in our example, the chain rule becomes invaluable. Each layer of the composition has its behavior described by its derivative:

• The outer function is influenced by the changes of the inner function.
• The derivative at any point depends on how both layers react to changes.

By understanding composite functions, we broaden our toolkit into a world where processes can be neatly broken down into simpler, manageable components. In essence, calculus provides the methods to dive deeper into these processes, revealing intricate details hidden within nested functions.Our work with \(\cos(\sin^{-1}(2w))\) illustrates these principles. It involves adjusting for both the nature of cosine and the special characteristics of arcsine, all scaled appropriately. By dissecting this composite function into its parts, we lay the groundwork to tackle even greater mathematical challenges.