Problem 26 - Calculus for Biology and Medicine

Question

In Problems \(1-58\), find the derivative with respect to the independent variable. $$ f(x)=-\cot \left(3 x^{3}-4 x\right) $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = -\csc^2(3x^3 - 4x)(9x^2 - 4) \).

Step 1: Identify the Function

We need to find the derivative of the function \( f(x) = -\cot(3x^3 - 4x) \). The function is a composition of the cotangent function with a polynomial, so we will apply the chain rule to find its derivative.

Step 2: Apply the Chain Rule

The chain rule states \( \frac{{d}}{{dx}}[f(g(x))] = f'(g(x)) \cdot g'(x) \). For our function, \( f(x) = -\cot(u) \) where \( u = 3x^3 - 4x \). First, find the derivative of \( -\cot(u) \), which is \( \csc^2(u) \), and then multiply by the derivative of \( u \).

Step 3: Differentiate the Inner Function

Differentiate the inner function \( u(x) = 3x^3 - 4x \). The derivative is \( u'(x) = 9x^2 - 4 \).

Step 4: Combine the Derivatives

Using the chain rule, the derivative is \( f'(x) = -\csc^2(u) \cdot u'(x) \). Substitute \( u = 3x^3 - 4x \) and \( u'(x) = 9x^2 - 4 \), giving:\[ f'(x) = -\csc^2(3x^3 - 4x) \cdot (9x^2 - 4) \].

Step 5: Simplify the Expression

The derivative of the function is \( f'(x) = -\csc^2(3x^3 - 4x)(9x^2 - 4) \). Since there's no further simplification possible, this is our final answer.

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental tool in calculus for finding the derivative of a composite function. Composite functions are essentially functions within functions, like layers of an onion.
Consider the function given in the exercise: \( f(x) = -\cot(3x^3 - 4x) \). Here, you have an outer function, which is the cotangent, and an inner function, \( u = 3x^3 - 4x \).
To differentiate a composite function like this one, you must:

• First differentiate the outer function \(-\cot(u)\). The derivative of \(\cot(u)\) is \(-\csc^2(u)\), so the derivative with respect to \(u\) would be \(\csc^2(u)\).
• Then, multiply this result by the derivative of the inner function \(u(x) = 3x^3 - 4x\), which is \(9x^2 - 4\).

In essence, the chain rule looks like a relay race where you first differentiate the outer runner, and then pass the baton to the inner runner, multiply their derivatives, and combine them to find the complete derivative.

Exploring Trigonometric Functions

Trigonometric functions like \( \sin \), \( \cos \), and \( \cot \) are periodic functions commonly used in calculus. The function \( \cot \) represents the cotangent, which is the reciprocal of the tangent function: \( \cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x} \).
In calculus, these functions play a crucial role in problems involving waves, cycles, and anything with circular or repetitive motion. When differentiating trigonometric functions like \( \cot(x) \), remember:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(\cos(x)\) is \(-\sin(x)\).
• The derivative of \(\tan(x)\) is \(\sec^2(x)\).
• For \(\cot(x)\), its derivative is \(-\csc^2(x)\), where \(\csc(x) = \frac{1}{\sin(x)}\).

When dealing with these, be careful and always check your work to ensure you use the correct derivative formulas. These derivatives are often seen in conjunction with the chain rule, especially in composite functions.

Getting a Grasp of Calculus

Calculus is the mathematical study of continuous change and is divided into two main branches: differential calculus and integral calculus.
In differential calculus, such as in this exercise, the focus is on defining and finding the derivative, which allows us to understand the rate of change. This is akin to understanding the speed of a car at any given instant, not just its average speed over time.
With the help of calculus, we can solve problems involving:

• Motion and dynamics by finding velocity and acceleration.
• Growth rates in economics or populations.
• Understanding patterns of change in physics and engineering.

The core ideas are finding the derivative (rate of change) and the integral (the area under curves), which together form the backbone of calculus. Mastering calculus opens up a world of applications and empowers you to solve complex real-world problems.