Problem 20 - Calculus for Biology and Medicine

Question

Find the derivative with respect to the independent variable. $$ f(x)=-\sin ^{2}\left(2 x^{3}-1\right) $$

Step-by-Step Solution

Answer

The derivative is \( 12x^2\\sin(2x^3 - 1)\\cos(2x^3 - 1) \).

Step 1: Recognize the Chain Rule

The function \( f(x) = -\sin^2(2x^3 - 1) \) can be differentiated using the chain rule. This method is helpful when dealing with composite functions. Here, our function is composed of an outer function \( u^2 \) and inner function \( u = -\sin(u) \) where \( u = 2x^3 - 1 \).

Step 2: Differentiate the Outer Function

Differentiate the outer function \( -u^2 \) with respect to \( u \). The derivative of \( -u^2 \) with respect to \( u \) is: \(-2u\).

Step 3: Differentiate the Inner Function

Consider the inner function \( \sin(2x^3 - 1) \). The derivative of \( -\sin(u) \) with respect to \( u \) is \( -\cos(u) \).

Step 4: Differentiate the Innermost Function

Differentiate the innermost function \( 2x^3 - 1 \) with respect to \( x \). The derivative is \( 6x^2 \).

Step 5: Apply the Chain Rule

Combine all these derivatives using the chain rule: \( \frac{d}{dx}[-\sin^2(2x^3 - 1)] = -2(-\sin(2x^3 - 1)) (-\cos(2x^3 - 1))(6x^2) \).

Step 6: Simplify the Expression

Simplify the derivative expression: \( -2(-\sin(2x^3 - 1)) (-\cos(2x^3 - 1))(6x^2) = 12x^2\sin(2x^3 - 1)\cos(2x^3 - 1) \).

Key Concepts

Derivatives

In calculus, derivatives represent the rate at which a function is changing at any given point. Essentially, they're the foundation of differential calculus. The derivative of a function provides the slope of the tangent line to the graph of the function at any specified point. This idea is crucial when analyzing the behavior of functions and their graphs.

When we say we are finding the derivative of a function like \( f(x) \), it involves calculating how \( f(x) \) changes as \( x \) changes. For polynomials, the process is straightforward. For instance, the derivative of \( x^n \) is \( nx^{n-1} \). However, when functions become more complex, as in the function from the exercise, which involves trigonometric and power functions, we need more advanced techniques, like the chain rule, to find derivatives.

Understanding derivatives is not just limited to differentiating one function from another; it's about understanding change and predicting future behavior of mathematical models.

Chain Rule

The chain rule is a fundamental rule in calculus, used to differentiate composite functions. Composite functions are those where one function

is applied to the result of another function. Think of it like a layered sandwich, and we must take each layer separately to eat it all!

If you have a function \( f(g(x)) \), where \( f \) and \( g \) are different functions, the chain rule states that the derivative of \( f(g(x)) \) with respect to \( x \) is found by multiplying the derivative of the outer function \( f \) and the derivative of the inner function \( g \). Mathematically, this is expressed as:

\[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]
In our exercise, we applied this rule multiple times:

• First, differentiating the outer function \(-u^2\).
• Then, moving to the sine function within the problem.
• Finally, differentiating the innermost polynomial function \(2x^3 - 1\).

The chain rule essentially tells us how to "chain" the derivative of the functions together correctly to get the complete answer.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are periodic functions widely used in calculus. They model scenarios with waves, cycles, and oscillations — think tides and sound waves.

Understanding how to differentiate trigonometric functions is essential. The derivatives of sine and cosine have simple forms:

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

In the given problem, differentiating involves applying these basic derivatives and recognizing their signs when combined with other functions.

Differentiating expressions involving trigonometric functions becomes more complex when they are nested within other functions, as in our example. Recognizing this nesting and using the chain rule efficiently are key skills.

Each step - from identifying trigonometric functions to applying the chain rule - leads to a detailed understanding of how these functions behave and how we can accurately measure their rate of change.