Problem 34 - Calculus for Biology and Medicine

Question

Find the derivative with respect to the independent variable. $$ f(x)=\frac{\cos (2 x)}{\tan (4 x)} $$

Step-by-Step Solution

Answer

The derivative of the function is \( f'(x) = \frac{-2\sin(2x)\tan(4x) - 4\cos(2x)\sec^2(4x)}{\tan^2(4x)} \).

Step 1: Identify the components

The function given is a quotient of two functions: numerator, \( u(x) = \cos(2x) \), and denominator, \( v(x) = \tan(4x) \). In order to differentiate the function \( f(x) = \frac{u(x)}{v(x)} \), the quotient rule will be used. The quotient rule states that if you have \( f(x) = \frac{u(x)}{v(x)} \), then \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \).

Step 2: Differentiate the numerator

Differentiate \( u(x) = \cos(2x) \) to find \( u'(x) \). The derivative of \( \cos(2x) \) is \( -2\sin(2x) \) using the chain rule. Hence, \( u'(x) = -2\sin(2x) \).

Step 3: Differentiate the denominator

Differentiate \( v(x) = \tan(4x) \) to find \( v'(x) \). The derivative of \( \tan(4x) \) is \( 4\sec^2(4x) \) using the chain rule. Hence, \( v'(x) = 4\sec^2(4x) \).

Step 4: Apply the quotient rule

Substitute \( u(x), u'(x), v(x), \) and \( v'(x) \) into the quotient rule formula. Therefore, \[ f'(x) = \frac{(-2\sin(2x))\tan(4x) - (\cos(2x))(4\sec^2(4x))}{\tan^2(4x)}. \]

Step 5: Simplify the expression

Now we simplify the expression obtained from Step 4. The expression for the derivative is \[ f'(x) = \frac{-2\sin(2x)\tan(4x) - 4\cos(2x)\sec^2(4x)}{\tan^2(4x)}. \] This can be left in this factorized form, or you can simplify further if needed to ..., but it can get quite complex beyond this point.

Key Concepts

Mastering the Quotient Rule: Differentiation of Fractions

The Quotient Rule is a powerful tool in calculus used to differentiate quotients of functions, like the one in our exercise. In simple terms, a quotient means one function divided by another. Here, we have two functions where the numerator is \( u(x) = \cos(2x) \) and the denominator is \( v(x) = \tan(4x) \). To find the derivative of such expressions, we need the quotient rule.

The Quotient Rule formula is: \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \] This formula helps to consider both the derivative of the top function \(u(x)\), and the derivative of the bottom function \(v(x)\), contributing to the overall change.

• Start by finding the derivatives of both functions separately.
• Then apply the formula by substituting these derivatives into it.
• Keep the denominator squared, as indicated in the formula.

Understanding this rule is fundamental when dealing with fractions in calculus, providing clarity and precision in differentiation.

Unraveling the Chain Rule: A Deep Dive

The Chain Rule is indispensable when dealing with composite functions, such as \( \cos(2x) \) and \( \tan(4x) \) in our example. It allows for the differentiation of functions within functions. If you see an operation within another, think of the chain rule.

The Chain Rule can be expressed as:

• \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \)

What this means is, you first differentiate the outer function, which in the exercise involves trigonometric expressions, and then multiply by the derivative of the inner function.

• For \( \cos(2x)\), the inner function is \(2x\). Differentiate \(\cos\) to get \(-\sin\), multipled by the derivative of \(2x\), giving us \( -2\sin(2x) \).
• For \( \tan(4x)\), the derivative of \( \tan(y)\) is \( \sec^2(y) \), and for the inner \(4x\), the derivative is \(4\). Combine to get \( 4\sec^2(4x) \).

By fully grasping the chain rule, you can solve increasingly complex calculus problems with functions inside functions.

Trigonometric Functions in Calculus: Unlocking their Secrets

Trigonometric functions like sine, cosine, and tangent are foundational in calculus. They frequently appear in differentiation problems due to their cyclical and repeating nature. Here's a quick look at how they operate in calculus.

Understanding derivatives of trigonometric functions is crucial:

• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• The derivative of \( \tan(x) \) is \( \sec^2(x) \).

These derivatives help in solving equations like the one in our example, where you have composite trigonometric expressions. The periodic nature of trigonometric derivatives can often simplify complex calculus problems.

When faced with \(\cos\) or \(\tan\) in calculus, remember to check if chain or quotient rules apply, as these rules often work hand-in-hand with trig derivatives. With practice, solving these becomes intuitive.