Problem 17 - Calculus for Biology and Medicine

Question

In Problems \(1-58\), find the derivative with respect to the independent variable. $$ f(x)=\sin ^{3}\left(x^{2}-3\right) $$

Step-by-Step Solution

Answer

The derivative is \( 6x \sin^2(x^2 - 3) \cos(x^2 - 3) \).

Step 1: Identify the Composition

The function given is \( f(x) = \sin^3(x^2 - 3) \). We identify that this expression is a composition of several functions: the outermost function is a cube \( (u^3) \), where \( u = \sin(x^2 - 3) \), and within \( u \), we have the sine function \( \sin(v) \), where \( v = x^2 - 3 \).

Step 2: Apply the Chain Rule for Outer Function

To find \( \frac{d}{dx}[\sin^3(x^2 - 3)] \), apply the chain rule. Begin with the outermost function \( u^3 \). The derivative of \( u^3 \) with respect to \( u \) is \( 3u^2 \). Therefore, \[ \frac{d}{dx}[\sin^3(x^2 - 3)] = 3\sin^2(x^2 - 3) \cdot \frac{d}{dx}[\sin(x^2 - 3)]. \]

Step 3: Differentiate the Sine Function

Now differentiate \( \sin(v) = \sin(x^2 - 3) \). Using the derivative of the sine function, we find \( \frac{d}{dv}[\sin(v)] = \cos(v) \). Thus, \[ \frac{d}{dx}[\sin(x^2 - 3)] = \cos(x^2 - 3) \cdot \frac{d}{dx}[x^2 - 3]. \]

Step 4: Differentiate the Inner Function

Compute \( \frac{d}{dx}[x^2 - 3] \). The derivative of \( x^2 \) with respect to \( x \) is \( 2x \), and the derivative of \(-3\) is \( 0 \), as it is a constant. Therefore, \( \frac{d}{dx}[x^2 - 3] = 2x \).

Step 5: Combine All Parts

Substitute back the derivative results into each other. So, \[ \frac{d}{dx}[\sin(x^2 - 3)] = \cos(x^2 - 3) \cdot 2x \]. Substituting into Step 2, we have \[ \frac{d}{dx}[\sin^3(x^2 - 3)] = 3\sin^2(x^2 - 3) \cdot \cos(x^2 - 3) \cdot 2x. \] Simplifying, \[ \frac{d}{dx}[\sin^3(x^2 - 3)] = 6x \sin^2(x^2 - 3) \cos(x^2 - 3). \]

Key Concepts

Chain Rule

The chain rule is a vital tool in calculus for finding the derivative of composite functions. It helps us differentiate a function which is composed of other functions. Think of it like peeling layers of an onion.
In the context of our exercise, we started by recognizing that the given function is a composition. It involved an outer cubic function and an inner sine function. The chain rule tells us that to differentiate the composition, we differentiate the outer function first and then multiply it by the derivative of its inner function.

• First, differentiate the outer function, keeping the inner function intact.
• Next, find the derivative of the inner function.
• Finally, multiply these derivatives together.

This systematic approach makes the chain rule an awesome method to tackle derivatives of stacked functions. It's especially helpful with multiple layers, as we've seen, there can be inner and even more inner functions inside!

Composite Functions

Composite functions combine two or more functions into one. They appear everywhere in calculus and need specific techniques, like the chain rule, to differentiate.
Our original problem involved a composite function:

• The outermost part is a cube, \(\sin^3 (x^2 - 3)\).
• Next, the sine function is wrapping another function, \(\sin(x^2 - 3)\).
• Finally, there is the innermost quadratic function, \(x^2 - 3\).

Each layer serves a purpose and affects how we find the derivative. The order is crucial in solving it: we start from the outside layer and work inwards when applying derivatives. Recognizing such compositions makes it easier to systematically apply the differentiation rules, ensuring we don't miss any crucial details.

Trigonometric Derivatives

Trigonometric derivatives are specific rules for differentiating trigonometric functions like sine, cosine, etc. They come in handy when these functions appear in compositions.
In our exercise, we had the sine function inside another function. Understanding how to differentiate trigonometric functions is essential.

• The derivative of the sine function is cosine, which we used directly to differentiate the sine part of our composite function.
• We multiplied it by the derivative of the innermost function, as the chain rule prescribes.

Building up from the simple rule \( \frac{d}{dx} [\sin(x)] = \cos(x) \) helps us handle more complex compositions that involve trigonometric functions. This underscores how each part of the derivative contributes to the whole, refining our skills in calculus and making trigonometric derivatives manageable.