Problem 7 - Single Variable Calculus

Question

Find the derivative of the function. $$ s=\sin x \cos x $$

Step-by-Step Solution

Answer

The derivative of the function \(s = \sin x \cos x\) is: \(s' = \cos^2 x - \sin^2 x\)

Step 1: Identify the functions

We have two functions in the given equation: 1. f(x) = \(\sin x\) 2. g(x) = \(\cos x\)

Step 2: Find the derivatives of the individual functions

Find the derivatives of f(x) and g(x) with respect to x: 1. f'(x) = \(\frac{d(\sin x)}{dx} = \cos x\) 2. g'(x) = \(\frac{d(\cos x)}{dx} = -\sin x\)

Step 3: Apply the product rule

According to the product rule, the derivative of the function s(x) = f(x) * g(x) is given by: s'(x) = f'(x)g(x) + f(x)g'(x) Now, substitute the derivatives of f(x) and g(x) we found, and the original functions into the equation: s'(x) = \((\cos x)(\cos x) + (\sin x)(-\sin x)\)

Step 4: Simplify the expression

Simplify the expression to get the final derivative: s'(x) = \(\cos^2 x - \sin^2 x\) The derivative of the function \(s = \sin x \cos x\) is: $$ s' = \cos^2 x - \sin^2 x $$

Key Concepts

Product Rule

The product rule is a fundamental principle used in calculus to find the derivative of the product of two functions. In simpler terms, if you have two functions, say \(f(x)\) and \(g(x)\), and you want to find the derivative of their multiplication, you can't just multiply their derivatives. Instead, you use the product rule, which states:
\[s'(x) = f'(x)g(x) + f(x)g'(x)\]
Here's how it works:

• Take the derivative of the first function \(f(x)\), which is \(f'(x)\), and multiply it by the second function \(g(x)\).
• Then, take the original first function \(f(x)\) and multiply it by the derivative of the second function \(g'(x)\).
• Finally, add these two results together to get the derivative of the entire product.

This rule is helpful each time you need to derive a function that is a multiplication of two or more functions. It's important because it preserves the interaction between the two functions during differentiation.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variable where both sides of the equation are defined. These identities simplify the process of solving problems involving angles and lengths in triangles. Some essential trigonometric identities include:

• Pythagorean Identity: \(\sin^2 x + \cos^2 x = 1\)
• Angle Sum and Difference Identities, like \(\sin(a \pm b) = \sin a \cos b \pm \cos a \sin b\)
• Double Angle Identities, such as \(\cos(2x) = \cos^2 x - \sin^2 x\), which is directly related to the derivative we found in the original problem.

By using these identities, complex trigonometric expressions can be simplified. This simplification is especially useful in calculus when calculating derivatives or integrals, as it can make a difficult problem more manageable. For instance, the derivative of \(s=\sin x \cos x\) simplifies to \(\cos^2 x - \sin^2 x\), which is directly a part of the double angle identity.

Sin and Cos Derivatives

The derivatives of the sine and cosine functions are among the most critical derivatives to understand in calculus. They are the building blocks for solving more complex problems involving trigonometric functions. Here is what you need to know:

• The derivative of \(\sin x\) with respect to \(x\) is \(\cos x\). This means that if you're differentiating a function and it includes \(\sin x\), its rate of change is given by \(\cos x\).
• The derivative of \(\cos x\) with respect to \(x\) is \(-\sin x\). This derivative indicates that for any changes in \(\cos x\), the rate is given by \(-\sin x\).

Remembering these derivatives helps significantly when applying the product rule to functions that involve trigonometry. In our worked example, correctly computing \(f'(x) = \cos x\) and \(g'(x) = -\sin x\) was essential for applying the product rule to find the derivative of \(s=\sin x \cos x\). These simple derivatives, when combined, unlock the complexity of various functions in calculus, proving that a solid foundation in basic derivatives is invaluable.