Problem 43 - Calculus Single Variable

Question

Calculate the derivative of each of the expressions in Exercises 39-44 by applying both the Product and Quotient Rules. $$ \sin ^{2}(x) \cdot \sec (x) $$

Step-by-Step Solution

Answer

The derivative is \( 2\sin(x)\cos(x)\sec(x) + \sin^2(x)\sec(x)\tan(x) \).

Step 1: Identify the Components for Product Rule

To apply the Product Rule, identify the two functions being multiplied. In this expression, let \( u(x) = \sin^2(x) \) and \( v(x) = \sec(x) \).

Step 2: Differentiate Each Function

Find the derivatives of \( u(x) \) and \( v(x) \). - The derivative of \( u(x) = \sin^2(x) \) is found by using the chain rule: \( \frac{du}{dx} = 2\sin(x)\cos(x) = \sin(2x) \).- The derivative of \( v(x) = \sec(x) \) is \( \frac{dv}{dx} = \sec(x)\tan(x) \).

Step 3: Apply the Product Rule

The Product Rule states: \( \frac{d}{dx}[u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x) \).- Substitute \( u(x) \), \( v(x) \), \( u'(x) \), and \( v'(x) \) into the Product Rule: \[\frac{d}{dx}[\sin^2(x) \cdot \sec(x)] = \sin(2x) \cdot \sec(x) + \sin^2(x) \cdot \sec(x)\tan(x)\]

Step 4: Simplify the Expression

Combine and simplify terms if necessary:- Recall that \( \sin(2x) = 2\sin(x)\cos(x) \) and simplify the expression:\[= 2\sin(x)\cos(x)\sec(x) + \sin^2(x)\sec(x)\tan(x)\]- Further simplification may combine terms within context, but this expression is sufficient for most purposes.

Key Concepts

Understanding Derivatives

Derivatives are a core concept in calculus, helping to determine how a function changes as its input changes. It's like finding the slope of a curve at any given point. For functions that involve multiple mathematical operations, like products or quotients, specific rules are used to find derivatives, such as the Product and Quotient Rules. These rules help in calculating the derivatives of complex expressions efficiently. For example, if you multiply two functions together, the Product Rule helps determine the rate of change of their product. When dealing with trigonometric functions, which we'll explore next, understanding how to derive them is particularly crucial.

Exploring Trigonometric Functions

Trigonometric functions like sine (\(\sin\)) and secant (\(\sec\)) play a fundamental role in calculus and numerous applications. In our example, we have \(\sin^2(x)\) and \(\sec(x)\). \(\sin^2(x)\) can be thought of as \((\sin(x))^2\), which means we are working with a function squared. \(\sec(x)\), the reciprocal of cosine (\(\cos(x)\)), becomes particularly important when differentiated due to its unique behavior at certain angles.
When differentiating trigonometric functions:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(\sec(x)\) is \(\sec(x)\tan(x)\).

Understanding these derivatives helps in applying the Product Rule correctly.

The Chain Rule: Why It's Important

The Chain Rule is indispensable when dealing with composite functions. It's used when a function is applied within another function, as seen with \(\sin^2(x)\). Here, \(\sin(x)\) is inside the square function, requiring the use of the Chain Rule to find its derivative. The chain rule states that if you have a function composed of \(f(g(x))\), then the derivative is \(f'(g(x))\cdot g'(x)\).
In our problem:

• Set \(f(u) = u^2\) and \(g(x) = \sin(x)\).
• Applying the Chain Rule, the derivative of \(\sin^2(x)\) is \(2\sin(x)\cdot\cos(x)\), or \(\sin(2x)\) using double angle resolution.

Mastering the Chain Rule alongside other rules allows you to tackle complicated calculus problems with ease.