Problem 48 - Calculus for Biology and Medicine

Question

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

Step 1: Identify the Functions

The given function is defined as the product of two trigonometric functions. Here, \( h(x) = \cot(3x) \csc(3x) \), which means we have a product of the functions \( g(x) = \cot(3x) \) and \( f(x) = \csc(3x) \). We need to apply the product rule for differentiation.

Step 2: Recall the Product Rule

The product rule states that if \( h(x) = u(x) \cdot v(x) \), then the derivative \( h'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \). We will let \( u(x) = \cot(3x) \) and \( v(x) = \csc(3x) \).

Step 3: Differentiate \( u(x) = \cot(3x) \)

The derivative of \( \cot(x) \) is \( -\csc^2(x) \). Therefore, the derivative of \( \cot(3x) \) with respect to \( x \) can be found using the chain rule: \( u'(x) = -\csc^2(3x) \cdot 3 = -3\csc^2(3x) \).

Step 4: Differentiate \( v(x) = \csc(3x) \)

The derivative of \( \csc(x) \) is \( -\csc(x) \cot(x) \). Thus, the derivative of \( \csc(3x) \) using the chain rule is \( v'(x) = -\csc(3x) \cot(3x) \cdot 3 = -3\csc(3x) \cot(3x) \).

Step 5: Apply the Product Rule

Using the product rule, plug the derivatives into the equation: \[h'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) = (-3\csc^2(3x)) \cdot \csc(3x) + \cot(3x) \cdot (-3\csc(3x)\cot(3x)).\]

Step 6: Simplify the Expression

Simplify each term: \[h'(x) = -3 \csc^3(3x) - 3 \cot^2(3x) \csc(3x).\]The simplified derivative of the function is:\[h'(x) = -3(\csc^3(3x) + \cot^2(3x) \csc(3x)).\]

Key Concepts

Product Rule

Differentiating products of functions is made simple with the product rule. Imagine two functions, say \( u(x) \) and \( v(x) \), that are multiplied together to get a new function \( h(x) = u(x) \cdot v(x) \). The product rule comes to the rescue when you need to find the derivative of \( h(x) \). It states: \[ h'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \] This rule applies the idea that you need to differentiate each function while considering how they interact with each other through multiplication. It helps to systematically break down and keep track of the individual transformations happening in the function. In our example, \( h(x) = \cot(3x) \csc(3x) \), and is composed of two trigonometric parts, necessitating the use of the product rule.

Trigonometric Functions

Trigonometric functions frequently appear in calculus, specifically when dealing with angles and periodic phenomena. Functions like \( \cot(x) \) and \( \csc(x) \) are less commonly introduced first, compared to \( \sin(x) \) and \( \cos(x) \), but are equally important.

• The cotangent function, \( \cot(x) \), is the reciprocal of the tangent function. It's expressed as \( \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)} \).
• The cosecant function, \( \csc(x) \), is the reciprocal of the sine function: \( \frac{1}{\sin(x)} \).

When differentiating, it's helpful to remember the derivatives of these functions:

• \( \frac{d}{dx}[\cot(x)] = -\csc^2(x) \)
• \( \frac{d}{dx}[\csc(x)] = -\csc(x) \cot(x) \)

Being familiar with these identities and derivatives simplifies the process when you encounter them individually or in combination within wider functions.

Chain Rule

The chain rule is crucial whenever you deal with composed functions – that is, functions within another function. It allows us to differentiate composite functions by focusing on the rates of change within each layer of the function. For a composite function \( f(g(x)) \), the derivative is: \[ \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \] Here, the derivative of the outer function \( f \) is taken with respect to its inside function \( g(x) \), then multiplied by the derivative of \( g(x) \) itself. In practice, consider \( \cot(3x) \) from our example: it's \( \cot(u) \) where \( u = 3x \). Calculating its derivative using the chain rule involves two steps:

• Determine the derivative of the outer function: \( \frac{d}{du}[\cot(u)] = -\csc^2(u) \).
• Multiply by the inner function's derivative: \( \frac{d}{dx}[3x] = 3 \).

Thus, \( \frac{d}{dx}[\cot(3x)] = -3 \csc^2(3x) \). Using the chain rule ensures accurate differentiation of complex functions.