Problem 57 - Calculus for Biology and Medicine

Question

Find the derivative with respect to the independent variable. $$ f(x)=\frac{\cos x^{2}}{\cos ^{2} x} $$

Step-by-Step Solution

Answer

The derivative is \(-2 \sin(x) (x \cos^{-1}(x) \sin(x^2) + \cos(x^2))\).

Step 1: Rewrite the Function

Before finding the derivative, rewrite the expression to make it easier to differentiate. We have the function:\[f(x) = \frac{\cos(x^2)}{\cos^2(x)}\]This can be rewritten as a product of functions using negative exponents:\[f(x) = \cos(x^2) \cdot \cos^{-2}(x)\]

Step 2: Apply the Product Rule

To find the derivative of a product of two functions, use the product rule, which states \[(uv)' = u'v + uv'\] For this function, let \( u = \cos(x^2) \) and \( v = \cos^{-2}(x) \). Now, differentiate each part separately: \[u' = \frac{d}{dx}[\cos(x^2)]\] and \[v' = \frac{d}{dx}[\cos^{-2}(x)]\]

Step 3: Differentiate \( u = \cos(x^2) \)

Use the chain rule to differentiate \( u = \cos(x^2) \):\[u' = \frac{d}{dx}[\cos(x^2)] = -\sin(x^2) \cdot \frac{d}{dx}[x^2] = -\sin(x^2) \cdot 2x = -2x\sin(x^2)\]

Step 4: Differentiate \( v = \cos^{-2}(x) \)

Use the chain rule to differentiate \( v = \cos^{-2}(x) \):\[v' = \frac{d}{dx}[\cos^{-2}(x)] = -2\cos^{-3}(x) \cdot \sin(x) = -2\sin(x)\cos^{-3}(x)\]

Step 5: Combine Using the Product Rule

Substitute the derivatives \( u' \) and \( v' \) into the product rule formula:\[(uv)' = (-2x\sin(x^2)) \cdot \cos^{-2}(x) + \cos(x^2) \cdot (-2\sin(x)\cos^{-3}(x))\]This simplifies to:\[f'(x)= -2x\sin(x^2)\cos^{-2}(x) - 2\cos(x^2)\sin(x)\cos^{-3}(x)\]

Step 6: Simplify the Expression

Factor and simplify the expression:First, take a common factor of \(-2\sin(x)\cos^{-3}(x)\):\[f'(x) = -2[ x\sin(x^2)\cos(x) + \cos(x^2) \sin(x) \cos^{-2}(x)]\]Finally, the simplified expression for the derivative is:\[f'(x) = -2 \sin(x) \left( x \cos^{-1}(x) \sin(x^2) + \cos(x^2) \right)\]

Key Concepts

Product Rule

When you need to find the derivative of a function that's a product of two distinct functions, the product rule is your go-to tool. This rule simplifies the process, allowing you to handle complex expressions more efficiently. The product rule formula is

• \( (uv)' = u'v + uv' \)

where \( u \) and \( v \) are functions of \( x \). In applying the product rule, ensure you first differentiate each function separately.
For instance, if \( u = \cos(x^2) \) and \( v = \cos^{-2}(x) \), you'll need to first find \( u' \) and \( v' \), the derivatives of \( u \) and \( v \) respectively. This means:

• Calculate \( u' = \frac{d}{dx}[\cos(x^2)] \)
• Calculate \( v' = \frac{d}{dx}[\cos^{-2}(x)] \)

Once you have those derivatives, insert them back into the product rule formula to find the derivative of the entire expression. Combining the results gives the derivative of the product, making it easier to handle multivariable expressions. Utilizing the product rule correctly is crucial in differential calculus and helps unravel many complex calculus problems.

Chain Rule

The chain rule is a powerful differentiation technique that's essential when dealing with composite functions. A composite function is essentially a function within a function. The chain rule comes into play by linking the rates of change of these nested functions. The rule

• is expressed as \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

This means you first differentiate the outer function, treating the inner one as a constant, and then multiply it by the derivative of the inner function.
An example of this is when differentiating \( u = \cos(x^2) \). Here, the function is \( \cos \) and the inner function is \( x^2 \). Applying the chain rule, we differentiate \( \cos \) to get \( -\sin(x^2) \) and multiply by the derivative of \( x^2 \), which is \( 2x \).
Thus,

• \( u' = -2x\sin(x^2) \)

The chain rule, much like the product rule, is a staple in calculus, enabling the differentiation of layered functions seamlessly.

Trigonometric Functions

Trigonometric functions like \( \sin(x) \), \( \cos(x) \), and their inverses, are foundational in calculus due to their periodic nature and frequent applications. When differentiating trigonometric expressions, knowledge of specific derivatives such as

• \( \frac{d}{dx}[\sin(x)] = \cos(x) \)
• \( \frac{d}{dx}[\cos(x)] = -\sin(x) \)

is essential.
The derivative of \( \cos(x) \) is particularly important in this exercise. More complex trigonometric expressions, such as \( \cos(x^2) \) and \( \cos^{-2}(x) \), can be dealt with by combining rules like the product rule and the chain rule. This combination allows us to navigate through multi-layered expressions involving trigonometric functions smoothly.
Remember, trigonometric derivatives often require simplification, especially when nested inside other functions. Being comfortable with trigonometric identities can further simplify differentiation and aid in deriving more manageable expressions.