Problem 44 - Calculus - AP Edition

Question

Find \(d y / d x\). $$y=\cos ^{-1}\left(\frac{x+1}{2}\right)$$

Step-by-Step Solution

Answer

\( \frac{d y}{d x} = -\frac{1}{\sqrt{4-(x+1)^2}} \)

Step 1: Identify the Outer Function

The given function is a composite function where \[ y = \cos^{-1}(u) \quad \text{and}\quad u = \frac{x+1}{2}. \] Here, \(y\) is the inverse cosine function applied to \(u\). We need to differentiate \(y\) with respect to \(u\) first, because the derivative of the inverse cosine, \(\cos^{-1}(x)\), is\[ \frac{d}{dx}[\cos^{-1}(x)] = -\frac{1}{\sqrt{1-x^2}}. \]

Step 2: Differentiate the Outer Function

Differentiating \(y = \cos^{-1}(u)\) with respect to \(u\) gives:\[ \frac{d y}{d u} = -\frac{1}{\sqrt{1-u^2}}. \]

Step 3: Differentiate the Inner Function

Since the inner function is \(u = \frac{x+1}{2}\), we differentiate \(u\) with respect to \(x\):\[ \frac{d u}{d x} = \frac{1}{2}. \]

Step 4: Apply the Chain Rule

Now, apply the chain rule which states \(\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}\).Substitute the derivatives found in Steps 2 and 3:\[ \frac{d y}{d x} = -\frac{1}{\sqrt{1-(\frac{x+1}{2})^2}} \cdot \frac{1}{2}. \]

Step 5: Simplify the Expression

Begin simplification of the expression inside the square root:\[ 1 - \left(\frac{x+1}{2}\right)^2 = 1 - \frac{(x+1)^2}{4}. \]Now further simplify:\[ 1 - \frac{(x+1)^2}{4} = \frac{4 - (x+1)^2}{4}. \]Therefore your expression for the derivative is:\[ \frac{d y}{d x} = -\frac{1}{2\sqrt{\frac{4 - (x+1)^2}{4}}} = -\frac{1}{\sqrt{4 - (x+1)^2}}. \]

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential tools in calculus for dealing with angles that produce a given trigonometric value. These functions include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \), among others. They allow us to find angles when we know the trigonometric value. For example, if \( \cos(C) = \frac{1}{2} \), then \( C = \cos^{-1}(\frac{1}{2}) \) tells us about the angle whose cosine is \( \frac{1}{2} \).
In calculus, understanding the derivatives of these functions is crucial:

• The derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1-x^2}} \).
• The derivative of \( \cos^{-1}(x) \) is \( -\frac{1}{\sqrt{1-x^2}} \).
• Note the negative sign in the derivative of the cosine inverse function, this is because the cosine function decreases over its domain of \([0, \pi]\).

It's important to recognize these derivatives when solving calculus problems involving inverse trigonometric functions.

Composite Functions

Composite functions are functions created when we combine two or more functions. For example, if we have \( f(x) \) and \( g(x) \), then the composite function \( f(g(x)) \) represents applying \( g(x)\) first and then applying \( f\) to the result.
In this exercise, \( y = \cos^{-1}(\frac{x+1}{2}) \), the function consists of the inverse cosine function applied to the expression \( \frac{x+1}{2} \). Here, you first evaluate the inner function \( u = \frac{x+1}{2} \), and then apply the cosine inverse to this result, producing \( y = \cos^{-1}(u) \).
When working with composite functions, it's important to:

• Identify the inner and outer functions correctly.
• Understand how to differentiate each function individually.

This understanding helps when applying the chain rule, especially in differentiating complex functions.

Chain Rule

The chain rule is a fundamental tool in calculus for finding derivatives of composite functions. It states that if you have two functions \( y = f(u) \) and \( u = g(x) \), the derivative \( \frac{dy}{dx} \) is found by multiplying the derivatives \( \frac{dy}{du} \) and \( \frac{du}{dx} \). This relationship is expressed in the formula: \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
Applying the chain rule allows you to break down complex differentiations into manageable parts. In the given example, to find \( \frac{dy}{dx} \) of \( y = \cos^{-1}(\frac{x+1}{2}) \):

• Differentiate the outer function: find \( \frac{dy}{du} = -\frac{1}{\sqrt{1-u^2}} \).
• Differentiate the inner function: find \( \frac{du}{dx} = \frac{1}{2} \).
• Multiply these derivatives: \( \frac{dy}{dx} = \left(-\frac{1}{\sqrt{1-(\frac{x+1}{2})^2}}\right) \times \frac{1}{2} \).

This process enables you to handle even the most complicated derivatives efficiently and correctly.