Problem 53 - Calculus - AP Edition

Question

Find \(d y / d x\). $$y=\sin ^{-1} x+\cos ^{-1} x$$

Step-by-Step Solution

Answer

The derivative is 0.

Step 1: Identify Known Trigonometric Identity

We know that \(\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\). This is a standard identity in trigonometry. Therefore, the function simplifies to \(y = \frac{\pi}{2}\).

Step 2: Calculate the Derivative of a Constant

Since \(y = \frac{\pi}{2}\) is a constant, its derivative with respect to \(x\) is 0. Therefore, \( \frac{d y}{d x} = 0 \).

Key Concepts

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the variables involved. These identities simplify the complex relationships between trigonometric functions into more manageable forms. One important identity that is frequently used in calculus is the sum of the inverse sine and inverse cosine functions: \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \). This identity implies that for any real number \( x \) within the domain \([-1, 1]\), the sum of the arcsine and arccosine of \( x \) is a constant, specifically \( \frac{\pi}{2} \).
This simplification helps break down problems that involve these inverse functions by reducing them to a constant value, thereby making differentiation straightforward.

Calculus Differentiation

Differentiation is a core concept in calculus used to find the rate of change of a function with respect to a variable. It is a fundamental technique used to solve problems involving various types of functions by finding their derivatives. In calculus, rules like the Power Rule, Product Rule, and Chain Rule help to simplify the process of differentiation for different functions.
When we talk about differentiating inverse trigonometric functions, we often use standard rules and identities. However, as shown in the original exercise, some trigonometric identities simplify the problem, and only basic differentiation is needed. The role of differentiation in this context is critical, as it allows us to see how a change in \( x \) affects \( y \). This is especially straightforward with constant functions (as in the given example) because their derivatives are zero, reflecting no rate of change.

Constant Function Differentiation

A constant function is one that does not change as the input varies. Mathematically, such a function is represented as \( y = c \), where \( c \) is a constant. When we differentiate a constant function with respect to any variable, such as \( x \), the result is always zero because a constant has no rate of change. This is expressed through the simple rule: \( \frac{d}{dx}(c) = 0 \).
In the context of the exercise provided, since \( y = \sin^{-1}x + \cos^{-1}x \) simplifies to \( y = \frac{\pi}{2} \), which is a fixed number, the differentiation becomes straightforward. This reflects the inherent beauty of constant function differentiation, showcasing calculus's power to handle even what appears as complex functions by reducing them to constants through the use of trigonometric identities.