Problem 48 - Calculus for Biology and Medicine

Question

In Problems \(1-58\), find the derivative with respect to the independent variable. $$ h(x)=\cot (3 x) \csc (3 x) $$

Step-by-Step Solution

Answer

The derivative is \( h'(x) = -3 \csc^3(3x)(\csc^2(3x) + \cot^2(3x)) \).

Step 1: Express the function

The function given is \( h(x) = \cot(3x) \csc(3x) \). We know that \( \cot(3x) = \frac{\cos(3x)}{\sin(3x)} \) and \( \csc(3x) = \frac{1}{\sin(3x)} \). Thus, \( h(x) = \frac{\cos(3x)}{\sin^2(3x)} \).

Step 2: Apply the Product Rule

Let \( u = \cot(3x) \) and \( v = \csc(3x) \). The derivative \( \frac{d}{dx}[uv] \) is \( u'v + uv' \). We need to find \( u' = \frac{d}{dx}[\cot(3x)] \) and \( v' = \frac{d}{dx}[\csc(3x)] \).

Step 3: Derivative of cotangent and cosecant

The derivative of \( \cot(3x) \) with respect to \( x \) is \( u' = -3 \csc^2(3x) \). The derivative of \( \csc(3x) \) with respect to \( x \) is \( v' = -3 \csc(3x)\cot(3x) \).

Step 4: Substitute the derivatives into the Product Rule

Substitute \( u', v \) and \( u, v' \) into the product rule formula: \[ h'(x) = (-3 \csc^2(3x)) \cdot \csc(3x) + \cot(3x) \cdot (-3 \csc(3x) \cot(3x)) \].

Step 5: Simplify the expression

The expression becomes \(-3 \csc^3(3x) - 3 \cot^2(3x) \csc(3x) \). Simplify to get the final result, \( h'(x) = -3 \csc(3x)(\csc^2(3x) + \cot^2(3x)) \). Use the identity \( \csc^2(x) = \cot^2(x) + 1 \) to further simplify: \( h'(x) = -3 \csc^3(3x) \sec^2(3x) \).

Key Concepts

Product Rule

In calculus, the product rule is a formula used to find the derivative of a product of two functions. It is extremely useful when dealing with expressions where two functions are multiplied together, just like in the scenario presented in the original exercise. The product rule states: if we have two functions, say \( u(x) \) and \( v(x) \), then the derivative of their product \( uv \) is given by:

• \( \frac{d}{dx}[uv] = u'v + uv' \)

Here \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively. In the given exercise, we identified our functions as \( u = \cot(3x) \) and \( v = \csc(3x) \). By applying the product rule, we can find the derivative of \( h(x) \). Whenever you encounter a similar problem, always break it down so you can clearly see which parts of the expression form \( u(x) \) and \( v(x) \). This makes it easier to apply the rule.

Trigonometric Functions

Trigonometric functions like sine, cosine, cotangent, and cosecant are fundamental in calculus and appear often in derivatives. These functions have specific rules that apply when differentiating them.

• \( \sin(x) \) and \( \cos(x) \) derivatives are basic: \( \frac{d}{dx}\sin(x) = \cos(x) \) and \( \frac{d}{dx}\cos(x) = -\sin(x) \).
• Cotangent \( \cot(x) \) and cosecant \( \csc(x) \) are less direct but are based on sine and cosine: \( \cot(x) = \frac{\cos(x)}{\sin(x)} \) and \( \csc(x) = \frac{1}{\sin(x)} \).

Understanding these functions as ratios of sine and cosine helps when applying derivative rules, such as the quotient or product rule, because it makes it easier to manipulate the expressions. In our exercise, substituting these ratios into \( h(x) \) was a crucial step before using the product rule.

Chain Rule

The chain rule is another powerful calculus rule that allows us to differentiate composite functions. Composite functions are where one function is nested inside another—like \( \cot(3x) \) or \( \csc(3x) \) in our case.

• The chain rule states that if you have a composition of functions \( f(g(x)) \), its derivative is \( f'(g(x)) \cdot g'(x) \).

In the problem, we applied the chain rule to find the derivatives of \( \cot(3x) \) and \( \csc(3x) \), taking into account the inner function \( 3x \). You'll notice that the derivative process involved multiplying by the derivative of the inner function \( 3 \), which is a key part of using the chain rule effectively.

Cosecant and Cotangent Derivatives

The derivatives of \( \cot(x) \) and \( \csc(x) \) are not as intuitive as those of sine and cosine, but they follow specific patterns:

• The derivative of \( \cot(x) \) is \( -\csc^2(x) \).
• The derivative of \( \csc(x) \) is \( -\csc(x)\cot(x) \).

In our exercise, you saw how these derivatives were computed with respect to \( 3x \) and involved using both the product rule and the chain rule. Getting comfortable with these derivatives is essential for tackling calculus problems involving trigonometric functions as they frequently appear in physics, engineering, and mathematics. Remember to always follow through with simplifying your final result by applying trigonometric identities, as was done in showing \( h'(x) = -3 \csc^3(3x) \sec^2(3x) \).