Problem 20 - Calculus for Biology and Medicine

Question

In Problems \(1-58\), 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: Identify the Function

The given function is written as \( f(x) = -\sin^2(2x^3 - 1) \). This is a composite function because it involves a sine function and a polynomial function.

Step 2: Apply the Chain Rule

To differentiate \( f(x) = -\sin^2(u) \), where \( u = 2x^3 - 1 \), we apply the chain rule. The chain rule states that \((g(h(x)))' = g'(h(x)) \cdot h'(x)\). In this case, we need to find the derivative of \( -\sin^2(u) \) and then multiply that by the derivative of \( u \).

Step 3: Differentiate the Outer Function

The outer function is \( -\sin^2(u) \). The derivative of \( -\sin^2(u) \) with respect to \( u \) is \( -2\sin(u) \cdot \cos(u) \) because the derivative of \( \sin^2(u) \) is \( 2\sin(u) \cdot \cos(u) \) by applying the chain rule to \( \sin(u) \cdot \sin(u) \).

Step 4: Differentiate the Inner Function

The inner function is \( u = 2x^3 - 1 \). Its derivative with respect to \( x \) is \( u' = 6x^2 \) because the derivative of \( 2x^3 \) is \( 6x^2 \) and the derivative of \(-1\) is \( 0 \).

Step 5: Combine Using the Chain Rule

Combine the derivative of the outer function and the inner function using the chain rule: \[\frac{d}{dx}f(x) = -2\sin(u)\cos(u) \cdot u' = -2\sin(2x^3 - 1)\cos(2x^3 - 1) \cdot 6x^2\].

Step 6: Simplify the Expression

Simplify the expression by multiplying the constants and terms: \[\frac{d}{dx}f(x) = -12x^2 \sin(2x^3 - 1) \cos(2x^3 - 1)\]. The term \( \sin(2x^3 - 1) \cos(2x^3 - 1) \) is equivalent to \( \frac{1}{2}\sin(2(2x^3 - 1)) \), but we will keep it in the given form for this problem.

Key Concepts

Understanding Derivatives

A derivative represents a fundamental concept in calculus. It indicates how a function changes as its input changes. In simple terms, the derivative can be viewed as the 'slope' of a function at any given point. When we differentiate a function like

• \(-\sin^2(2x^3 - 1)\), we are finding how the function value shifts with tiny changes in \(x\).
• Derivatives can help us understand rates of change, such as speed, acceleration, or growth.

These are key applications in sciences and engineering, making the concept of derivatives both powerful and essential.

Composite Functions and the Chain Rule

A composite function is a function that is formed by combining two or more functions. In the expression \(-\sin^2(2x^3 - 1)\), you are dealing with a composite function.

• This composite structure is typical when you see one function inside another, such as a trigonometric function over a polynomial.
• To differentiate it, we employ the chain rule. The chain rule is a technique for calculating the derivative of such functions by taking the derivative of the outer function and multiplying it by the derivative of the inner function.

For example, using the chain rule on \(-\sin^2(u)\) involves first differentiating the outer function, \(-\sin^2\), and then the inner function, \(u=2x^3 - 1\). Both of these derivatives are stitched together by multiplication, giving us a complete derivative for the composite function.

Trigonometric Differentiation

Trigonometric differentiation is the application of derivative rules specifically for trigonometric functions like sine, cosine, and tangent. These functions have particular differentiation rules that students should memorize for ease of calculations.

• For instance, the derivative of \(\sin(x)\) is \(\cos(x)\), and this transforms when dealing with powers of sine.
• In \(-\sin^2(u)\), you apply the chain rule that engages both the derivative of \(\sin^2(u)\) and the trigonometric identity \(\sin(2A) = 2\sin(A)\cos(A)\).
• This yields \(-2\sin(u)\cos(u)\) for the derivative of the trigonometric part.

Trigonometric differentiations are quite prevalent in exercises involving oscillations, waves, and periodic phenomena, hence sharpening your skills in them is advantageous.