Problem 42 - Single Variable Calculus

Question

Find the derivative of the function. $$ y=\frac{x+\sin 2 x}{2+\cos 3 x} $$

Step-by-Step Solution

Answer

The derivative of the function \(y =\frac{x+\sin 2x}{2+\cos 3x}\) is given by: \(y' = \frac{2 + \cos 3x + 4\cos 2x + 2\cos 2x\cos 3x + 3x\sin 3x + 3\sin 2x\sin 3x}{(2+\cos 3x)^2}\)

Step 1: Identify u and v

In our given function, we have: $$ y=\frac{x+\sin 2x}{2+\cos 3x} $$ Let's identify u and v: - \(u(x) = x + \sin 2x\) - \(v(x) = 2 + \cos 3x\)

Step 2: Calculate the derivatives of u and v with respect to x

Now, we'll find the derivatives u'(x) and v'(x). For \(u'(x)\): $$ u'(x) = \frac{d}{dx}(x+\sin 2x) = 1 + \cos 2x\cdot 2 = 1 + 2\cos 2x $$ For \(v'(x)\): $$ v'(x) = \frac{d}{dx}(2+\cos 3x) = -\sin 3x\cdot 3 = -3\sin 3x $$

Step 3: Plug u, v, u', and v' into the quotient rule formula

Next, we'll apply the quotient rule formula with u, v, u' and v': $$ y' = \frac{u'\cdot v - u\cdot v'}{v^2} = \frac{(1 + 2\cos 2x)(2+\cos 3x) - (x+\sin 2x)(-3\sin 3x)}{(2+\cos 3x)^2} $$

Step 4: Simplify the obtained expression

Finally, let's simplify the expression of y' from step 3: $$ y' = \frac{2 + \cos 3x + 4\cos 2x + 2\cos 2x\cos 3x + 3x\sin 3x + 3\sin 2x\sin 3x}{(2+\cos 3x)^2} $$ So, the derivative of the given function is: $$ y' = \frac{2 + \cos 3x + 4\cos 2x + 2\cos 2x\cos 3x + 3x\sin 3x + 3\sin 2x\sin 3x}{(2+\cos 3x)^2} $$

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus. They represent the rate at which a function is changing at any given point. In simple terms, the derivative of a function gives you the slope of the tangent line at any point on the function's graph.
To find the derivative, you often use different rules depending on the form of the function. In this case, with a function written as a quotient, we use the quotient rule. The quotient rule is expressed as:

• If you have a function \( y = \frac{u(x)}{v(x)} \), then the derivative \( y' \) is: \( y' = \frac{u' \, v - u \, v'}{v^2} \).
• Here, \( u(x) \) and \( v(x) \) are functions of \( x \), and \( u'(x) \) and \( v'(x) \) are their derivatives.

Calculating derivatives with the quotient rule involves these steps:

• First, identify \( u(x) \) and \( v(x) \) from the function.
• Next, find the derivatives \( u'(x) \) and \( v'(x) \).
• Then apply these values to the quotient rule formula.

Trigonometric Functions

Trigonometric functions like sine and cosine play an important role in calculus, especially when finding derivatives. These functions model periodic phenomena, such as waves. When dealing with functions that involve these, you need to apply specific rules for differentiation.
For the sine function:- The derivative of \( \, \sin(x) \, \) is \( \, \cos(x) \, \).
For the cosine function:- The derivative of \( \, \cos(x) \, \) is \( \, -\sin(x) \, \).
When you have trigonometric functions with multipliers inside, such as \( \, \sin(2x) \, \) or \( \, \cos(3x) \, \), don't forget to use chain rule, which involves multiplying by the derivative of the inner function.
The example function \( y = \frac{x + \sin(2x)}{2 + \cos(3x)} \) requires you to calculate:

• \( \frac{d}{dx}(\sin 2x) = 2 \cos 2x \)
• \( \frac{d}{dx}(\cos 3x) = -3 \sin 3x \)

This chain rule application is crucial to reach the correct derivative of the function.

Simplification of Expressions

Simplifying expressions is a key step after calculating derivatives to ensure the final answer is in its simplest form. In our exercise, after applying the quotient rule, we arrived at a complex expression:\[ y' = \frac{(1 + 2\cos 2x)(2 + \cos 3x) - (x + \sin 2x)(-3\sin 3x)}{(2 + \cos 3x)^2} \]
To simplify:

• Start by expanding the elements in the numerator.
• Add and subtract like terms from the expanded expression.
• Check and ensure that the denominator remains unchanged.

Through simplification, we reduce the expression to a form that is easier to understand and work with. This leads to:\[ y' = \frac{2 + \cos 3x + 4\cos 2x + 2\cos 2x\cos 3x + 3x\sin 3x + 3\sin 2x\sin 3x}{(2 + \cos 3x)^2} \]True mastery of calculus includes developing the ability to simplify expressions effectively, allowing solutions to be neatly packaged and clearly communicated.