Problem 37 - Calculus for Biology and Medicine

Question

In Problems \(1-58\), find the derivative with respect to the independent variable. $$ f(x)=\sin (2 x-1) \cos (3 x+1) $$

Step-by-Step Solution

Answer

The derivative of the function is \( 2\cos(2x-1)\cos(3x+1) - 3\sin(2x-1)\sin(3x+1) \).

Step 1: Apply the Product Rule

The given function is the product of two functions: \( \sin(2x-1) \) and \( \cos(3x+1) \). The derivative of a product of two functions \( u(x) \) and \( v(x) \) is given by \( (uv)' = u'v + uv' \). Identify \( u(x) = \sin(2x-1) \) and \( v(x) = \cos(3x+1) \).

Step 2: Differentiate \( u(x) = \sin(2x-1) \)

Apply the chain rule to find the derivative of \( u(x) = \sin(2x-1) \). The derivative of \( \sin(u) \) is \( \cos(u) \cdot u' \). Here, \( u = 2x-1 \), so \( u' = 2 \). Thus, \( u'(x) = \cos(2x-1) \times 2 \).

Step 3: Differentiate \( v(x) = \cos(3x+1) \)

Apply the chain rule to differentiate \( v(x) = \cos(3x+1) \). The derivative of \( \cos(v) \) is \(-\sin(v) \cdot v' \). Here, \( v = 3x+1 \), so \( v' = 3 \). Thus, \( v'(x) = -\sin(3x+1) \times 3 \).

Step 4: Apply the derivatives to the Product Rule

Substitute the derivatives found into the product rule formula: \( (uv)' = u'v + uv' \). This results in: \( \cos(2x-1) \cdot 2 \cdot \cos(3x+1) + \sin(2x-1) \cdot (-\sin(3x+1) \cdot 3) \).

Step 5: Simplify the Expression

Simplify the expression to obtain the final derivative: \( 2\cos(2x-1)\cos(3x+1) - 3\sin(2x-1)\sin(3x+1) \).

Key Concepts

Product Rule

The product rule is an essential tool when dealing with derivatives of functions that are multiplied together. Imagine having two functions, say \( u(x) \) and \( v(x) \), and you want to find the derivative of their product. The product rule states that the derivative of \( u(x) \) times \( v(x) \) is given by:- \( (uv)' = u'v + uv' \).This means you first take the derivative of the first function \( u(x) \), and multiply it by the second function \( v(x) \). Then, you add the first function \( u(x) \) multiplied by the derivative of the second function \( v(x) \). Breaking it down into an easy-to-follow process:

• Differentiate the first function.
• Multiply by the unchanged second function.
• Add the product of the first function and the derivative of the second function.

For the exercise, apply the product rule to \( \sin(2x-1) \) and \( \cos(3x+1) \) to properly find the derivative.

Chain Rule

The chain rule helps us differentiate complex functions known as composite functions, which are functions within functions. Imagine having a function in the form \( f(g(x)) \). The chain rule states that the derivative is the outer function's derivative evaluated at the inner function and multiplied by the derivative of the inner function. In mathematical terms, if \( f(x) = h(g(x)) \), then:- \( f'(x) = h'(g(x)) \cdot g'(x) \).To put it into a simple method:

• Identify the inner function \( g(x) \).
• Differentiate it to find \( g'(x) \).
• Differentiate the outer function \( h \), treating the inner function as the variable, to get \( h'(g(x)) \).
• Multiply \( h'(g(x)) \) by \( g'(x) \).

In our exercise, you see the chain rule applied in differentiating \( \sin(2x-1) \) and \( \cos(3x+1) \). We focus on the inner functions \( 2x-1 \) and \( 3x+1 \), finding their derivatives first.

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in calculus. These are periodic functions and have well-known derivatives: - The derivative of \( \sin(x) \) is \( \cos(x) \).- The derivative of \( \cos(x) \) is \( -\sin(x) \).When you differentiate a trigonometric function, you are often dealing with these derivatives in conjunction with the chain rule or product rule.For example, in the exercise, we differentiated \( \sin(2x-1) \). - Using the fact that the derivative of \( \sin \) is \( \cos \) and applying the chain rule, we find it to be \( \cos(2x-1) \times 2 \).Similarly, for \( \cos(3x+1) \):- The derivative becomes \(-\sin(3x+1) \times 3\), after applying the chain rule.These principles are your guiding tools when tackling derivatives involving trigonometric functions.