Problem 12 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(7-12,\) use differentiation to verify the antiderivative formula. $$\int \frac{1}{\sqrt{1-u^{2}}} d u=\sin ^{-1} u+C$$

Step-by-Step Solution

Answer

Yes, the derivative of \(\sin^{-1}u + C\) is \(\frac{1}{\sqrt{1-u^2}}\), and this verifies the given antiderivative formula.

Step 1: Understand the task

First, it's important to remember that if \(F(u)\) is an antiderivative of \(f(u)\), then the derivative of \(F(u)\) should equal \(f(u)\). Thus if we can show that the derivative of \(\sin^{-1}u + C\), where C is constant, equals to \(\frac{1}{\sqrt{1-u^2}}\), we will have verified the given antiderivative formula.

Step 2: Differentiate the function

The derivative of \(\sin^{-1}u\) by the chain rule is \(\frac{1}{\sqrt{1-u^2}}\). And the derivative of any constant (in our case \(C\)) is zero. Hence, overall, the derivative of \(\sin^{-1}u + C\) is indeed \(\frac{1}{\sqrt{1-u^2}}\).

Key Concepts

Differentiation

Differentiation is like discovering the secret speed of functions. It tells us how fast a function is changing at any point. This can be done using derivatives, which are like special formulas that hunt down these rates of change.

When you differentiate a function, you're exploring its behavior by checking its slope or gradient. For instance, when you differentiate the function \( \sin^{-1} u + C \), you are looking at how this function changes as \( u \) changes. The result is given by the derivative \( \frac{1}{\sqrt{1-u^2}} \). Here, the constant \( C \) disappears because its rate of change is zero.

Differentiating inverse trigonometric functions like \( \sin^{-1} u \) requires specific rules due to their unique behavior compared to regular trigonometric functions. Using these rules effectively gives us the derivative needed to confirm an antiderivative formula.

Integration

Integration is the magical process of 'adding up' parts to find a whole. It is essentially the reverse operation of differentiation. While differentiation examines how functions change, integration attempts to reverse this, finding the original function from its rate of change.

The task of integration can be compared to assembling a puzzle. When you see an expression like \( \int \frac{1}{\sqrt{1-u^{2}}} \, du \), integration pieces it back into the familiar function \( \sin^{-1} u + C \). This happens because integration identifies the original function that matches the change expressed by the derivative.

Thinking about integration in terms of geometry, it can also calculate areas under curves. However, in calculus, the key is to understand how an antiderivative represents the cumulative effects of change embodied by the original function.

Inverse Trigonometric Functions

Inverse trigonometric functions turn angles back into ratios. They solve the problem of finding the angle, given a trigonometric ratio. For example, whereas \( \sin(u) = x \) finds \( x \) from \( u \), \( \sin^{-1}(x) \) confirms which angle \( u \) corresponds to a value \( x \).

These functions are crucial in calculus because they offer solutions to integrals that are not expressed using basic trigonometric or algebraic expressions. In our problem, \( \sin^{-1} u \) is an example of an inverse trigonometric function that we derive from the integral of \( \frac{1}{\sqrt{1-u^2}} \).

Dealing with inverse trigonometric functions sometimes requires careful handling of domain restrictions, but getting familiar with them enables solving a broader range of calculus problems, especially those involving integration and differentiation. They help translate real-world problems into manageable calculus expressions.