Problem 7 - Technical Mathematics with Calculus

Question

First Derivatives Find the derivative. $$y=\sin x \cos x$$

Step-by-Step Solution

Answer

\(y' = \(d/dx\) [\sin x \cdot \cos x] = \cos^2 x - \sin^2 x\)

Step 1: Identify the Function to Differentiate

Recognize that the function given is a product of two functions, which are \(y = \sin x \cdot \cos x\).

Step 2: Apply the Product Rule

Since the function is a product of two differentiable functions, use the product rule for differentiation, which is \( \(d/dx\) [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\). Let \(u(x) = \sin x\) and \(v(x) = \cos x\).

Step 3: Differentiate Each Function Individually

Differentiate \(u(x) = \sin x\) to get \(u'(x) = \cos x\), and differentiate \(v(x) = \cos x\) to get \(v'(x) = -\sin x\).

Step 4: Apply the Derivatives into the Product Rule

Plugging \(u'(x)\) and \(v'(x)\) into the product rule formula gives \( \(d/dx\) [\sin x \cdot \cos x] = \cos x \cdot \cos x + \sin x \cdot (-\sin x)\).

Step 5: Simplify the Expression

The derivative simplifies to \( \(d/dx\) [\sin x \cdot \cos x] = \cos^2 x - \sin^2 x\).

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus for finding the derivative of a product of two functions. Suppose you have functions, \(f(x)\) and \(g(x)\), that are both differentiable on an interval. According to the product rule, the derivative of their product \(h(x) = f(x)g(x)\) is given by \(h'(x) = f'(x)g(x) + f(x)g'(x)\). This means to find the derivative of the product, you take the derivative of the first function times the second function, and add this to the first function times the derivative of the second function.

In the given exercise, \(y = \text{sin} x \text{cos} x\) represents the product of two functions, \(\text{sin} x\) and \(\text{cos} x\). Applying the product rule simplifies the task of finding the derivative by breaking it down into more manageable steps. By differentiating each function separately and then applying the product rule formula, you can efficiently determine the derivative of the entire expression.

Differentiation Techniques

Differentiation is the process of finding the derivative, which measures how a function changes as its input changes. Differentiation techniques involve rules and methods to calculate the derivative of various types of functions efficiently. Some of these techniques include the product rule, quotient rule, chain rule, power rule, and differentiation of trigonometric, exponential, and logarithmic functions.

Mastering these techniques allows you to tackle a wide array of problems in calculus. For instance, once familiar with trigonometric functions and their derivatives, such as knowing that the derivative of \(\text{sin} x\) is \(\text{cos} x\) and the derivative of \(\text{cos} x\) is \(-\text{sin} x\), you can handle differentiation tasks involving these functions with ease. Skilled use of these techniques leads to swift and accurate computation of derivatives, which is essential for analyzing the behavior of functions.

Trigonometric Functions

Trigonometric functions, such as sine (\(\text{sin} x\)), cosine (\(\text{cos} x\)), and tangent (\(\text{tan} x\)), have significant importance in calculus due to their periodic properties and applications in modeling cyclical phenomena. These functions are defined using the angles and sides of a right-angled triangle, but their properties extend well beyond geometry into various fields of science and engineering.

In differentiation, it's particularly important to know the derivatives of these functions, which are cyclic themselves. The derivative of \(\text{sin} x\) is \(\text{cos} x\), while the derivative of \(\text{cos} x\) is \(-\text{sin} x\). This knowledge is applied in the product rule, as seen in the exercise, to differentiate products involving trigonometric functions. Understanding the behavior and derivatives of trigonometric functions is indispensable for solving calculus problems related to wave motion, signal processing, and oscillations, to name a few.