Problem 70 - Calculus for Biology and Medicine

Question

Find the derivative of $$ f(x)=\cos ^{2}\left(2 x^{2}+3\right) $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = -8x \cdot \cos(2x^2 + 3) \cdot \sin(2x^2 + 3). \)

Step 1: Identify the Functions Involved

To find the derivative of the function \( f(x) = \cos^2(2x^2 + 3) \), we first recognize that this is a composite function. It is a combination of the function \( g(u) = u^2 \) where \( u = \cos(v) \), and \( v = 2x^2 + 3 \).

Step 2: Apply the Chain Rule for Two Levels

The function is \( f(x) = (\cos(2x^2 + 3))^2 \). According to the chain rule, the derivative \( f'(x) \) is given by the derivative of the outer function times the derivative of the inner function:\[ f'(x) = 2 \cdot \cos(2x^2 + 3) \cdot \frac{d}{dx} \left( \cos(2x^2 + 3) \right). \]

Step 3: Differentiate the Innermost Function

The inner function to differentiate next is \( \cos(2x^2 + 3) \). The derivative is:\[ \frac{d}{dx}(\cos(2x^2 + 3)) = -\sin(2x^2 + 3) \cdot \frac{d}{dx}(2x^2 + 3). \]

Step 4: Differentiate the Innermost Quadratic Function

The innermost part is \( 2x^2 + 3 \). Differentiate this to obtain:\[ \frac{d}{dx}(2x^2 + 3) = 4x. \]

Step 5: Combine All the Derivatives

Substitute the previous derivatives:- From Step 3: \(-\sin(2x^2 + 3) \cdot 4x.\)- From Step 2, plug into the derivative:Combining, the derivative of \( f(x) \) is:\[ f'(x) = 2 \cdot \cos(2x^2 + 3) \cdot (-\sin(2x^2 + 3) \cdot 4x). \]

Step 6: Simplify the Expression

Simplify the expression to get:\[ f'(x) = -8x \cdot \cos(2x^2 + 3) \cdot \sin(2x^2 + 3). \]

Key Concepts

Chain Rule

The Chain Rule is a fundamental technique in calculus for finding the derivative of a composite function. Whenever you have a function composed of other functions, this rule helps us take it apart and differentiate it step-by-step. In essence, the Chain Rule states that to differentiate a composite function, you multiply the derivative of the outer function by the derivative of the inner function. This process entails:

• Identifying the outer and inner functions.
• Taking the derivative of the outer function while keeping the inner function unchanged.
• Multiplying this derivative by the derivative of the inner function.

For our example, the Chain Rule is applied to the function \((\cos(2x^2 + 3))^2\). By differentiating the squared cosine function first, and then the cosine itself, and finally the quadratic expression \(2x^2 + 3\), we piece together all parts to compute the final derivative.

Composite Function

In our problem, the function \(f(x) = \cos^2(2x^2 + 3)\) is a classic example of a composite function. Composite functions are formed when one function is applied to the result of another function. This layering makes them intriguing and also challenging.Understanding composite functions requires us to identify the separate functions that make up the whole. In our case:

• We see the outer function as \(g(u) = u^2\), where \(u = \cos(v)\).
• The middle layer is the trigonometric component \(\cos(v)\), where \(v = 2x^2 + 3\).
• Finally, the deepest function is the quadratic \(v = 2x^2 + 3\).

Breaking down the function in this way simplifies the differentiation process. We can use the Chain Rule to find the derivative by taking each function layer by layer, making each step manageable.

Trigonometric Function

Trigonometric functions, like cosine, sine, and tangent, frequently appear in calculus problems. In this particular exercise, the cosine function is an integral part of the composite function we need to differentiate. The differentiation rules for trigonometric functions are fundamental building blocks in calculus.Here’s how they relate to our exercise:

• The function \(\cos(2x^2 + 3)\) needs to be differentiated as part of finding \(f'(x)\).
• Using the derivative rule for cosine, \(\frac{d}{dx}\, \cos(u) = -\sin(u) \cdot \frac{du}{dx}\), you reveal how the trigonometric function reacts to changes in \(x\).
• Once the inside argument \(2x^2 + 3\) is differentiated to \(4x\), it further modifies the derivative result.

This interplay of trigonometric rules and the Chain Rule enables us to correctly derive complex derivatives like the one in this problem.