Problem 69 - Calculus for Biology and Medicine

Question

Find the derivatives of the following functions: $$ f(x)=\sin ^{2}\left(x^{2}-1\right) $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = 2x \sin(2(x^2 - 1)) \).

Step 1: Recognize the Composition

Identify that the function is a composition of functions. Here, \( f(x) = \sin^2(x^2-1) \) can be rewritten as \( f(x) = [\sin(g(x))]^2 \), where \( g(x) = x^2 - 1 \).

Step 2: Apply Chain Rule

We have a composition of functions, so we use the chain rule. The outer function is \( u^2 \), where \( u = \sin(g(x)) \), and the inner function is \( g(x) = x^2 - 1 \).

Step 3: Differentiate the Outer Function

Compute the derivative of the outer function with respect to \( u \): \( \frac{d}{du} (u^2) = 2u \). Substitute back \( u = \sin(x^2 - 1) \) to get \( 2 \sin(x^2 - 1) \).

Step 4: Differentiate the Inner Function

Compute the derivative of \( u = \sin(g(x)) \) with respect to \( x \). Use the chain rule again: \( \frac{d}{dx} \sin(g(x)) = \cos(g(x)) \cdot g'(x) \), where \( g(x) = x^2 - 1 \) and \( g'(x) = 2x \).

Step 5: Combine Derivatives

Combine the derivatives from Step 3 and Step 4. The derivative of \( f(x) \) is:\[ f'(x) = 2\sin(x^2 - 1) \cdot \cos(x^2 - 1) \cdot 2x \]Simplify it to:\[ f'(x) = 4x \sin(x^2 - 1) \cos(x^2 - 1) \].

Step 6: Use Trigonometric Identity

Recognize the identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \). Thus, \[ f'(x) = 2x \sin(2(x^2 - 1)) \].

Key Concepts

Derivatives

In calculus, a derivative represents the rate at which a function is changing at any given point. It is a core concept that measures how a particular function changes as its input changes. For example, in the function\( f(x) = \sin^2(x^2-1) \), the derivative helps us find the rate of change of this function with respect to \( x \).
To derive \( f(x) \), we need to figure out the function's dependency on the variable \( x \) using the rules of differentiation.

• The basic rule is that for \( x^n \), the derivative \( \frac{d}{dx}(x^n) = nx^{n-1} \).
• For trigonometric functions like sine, \( \frac{d}{dx}(\sin x) = \cos x \).

This foundation is critical because it connects us to more complex rules like the chain rule, which is essential for composing functions.

Chain Rule

The chain rule is a vital technique in calculus used to find the derivative of composite functions. Composite functions are essentially functions within functions, such as\( f(x) = \sin^2(x^2-1) \).
For such functions, we apply the chain rule which says that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

• This involves identifying the outer and inner functions. In our example, \( f(x) = [\sin(g(x))]^2 \), the outer function is \( u^2 \) where \( u = \sin(g(x)) \).
• The inner function here is \( g(x) = x^2 - 1 \).

The chain rule guides us through differentiating complex functions step by step, by handling each layer of the function individually.
Therefore, this rule allows us to unpack these layers mathematically, helping to simplify what seems complicated at first sight.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the included variables. One crucial identity is \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \).
This identity is particularly useful when simplifying derivatives that involve trigonometric functions. In the derivative of \( f(x) = \sin^2(x^2-1) \), we used the identity:\[ \sin(2(x^2-1)) = 2\sin(x^2-1)\cos(x^2-1) \]

• By recognizing such identities, we can express derivatives in simpler, more manageable forms, such as reducing the complexity of \( 2\sin(x^2-1)\cos(x^2-1) \).
• This simplification is often the last step in solving derivative problems, providing a cleaner and more insightful result.

Utilizing these identities not only helps in calculus, but is essential across many areas of mathematics and engineering.