Problem 23 - Single Variable Calculus

Question

Find the derivative of each function. \(f(x)=\frac{x}{x+2}\)

Step-by-Step Solution

Answer

The derivative of \(f(x) = \frac{x}{x+2}\) is \(f'(x) = \frac{2}{(x+2)^2}\).

Step 1: Identify u(x) and v(x)

Here, we are given the function \(f(x) = \frac{x}{x+2}\). We can identify \(u(x) = x\) and \(v(x) = x+2\).

Step 2: Find the derivatives of u(x) and v(x)

Now, we need to find the derivatives of \(u(x)\) and \(v(x)\). For \(u(x) = x\), the derivative is: \(u'(x) = \frac{d(u(x))}{dx} = \frac{d(x)}{dx} = 1\) For \(v(x) = x + 2\), the derivative is: \(v'(x) = \frac{d(v(x))}{dx} = \frac{d(x+2)}{dx} = 1\)

Step 3: Apply the Quotient Rule to find the derivative of f(x)

Now that we have the derivatives of \(u(x)\) and \(v(x)\), we can use the Quotient Rule to find the derivative of \(f(x)\). Recall the Quotient Rule for derivatives: \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \) Applying this rule to \(f(x) = \frac{x}{x+2}\), we get: \(f'(x) = \frac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}\) Substitute the known values of \(u(x)\), \(v(x)\), \(u'(x)\), and \(v'(x)\) into the equation: \(f'(x) = \frac{(1)(x+2)-(x)(1)}{(x+2)^2}\)

Step 4: Simplify the derivative of f(x)

Now, we will simplify the numerator and put it in a simplified form. \(f'(x) = \frac{x+2-x}{(x+2)^2}\) This simplifies to: \(f'(x) = \frac{2}{(x+2)^2}\) So, the derivative of the given function is: \(f'(x) = \frac{2}{(x+2)^2}\).

Key Concepts

Quotient Rule

The Quotient Rule is a fantastic tool in calculus for finding the derivative of a function that is expressed as a ratio of two other functions. If you have a function in the form of \( \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are functions differentiable in terms of \( x \), the Quotient Rule comes to your rescue.
The formula for the Quotient Rule is:
\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \] In simpler terms, it involves differentiating the numerator \( u(x) \) and denominator \( v(x) \) separately.

• Find \( u'(x) \) by differentiating \( u(x) \)
• Find \( v'(x) \) by differentiating \( v(x) \)
• Plug \( u'(x) \), \( v'(x) \), \( u(x) \), and \( v(x) \) back into the formula

This method ensures that you can correctly derive the rate of change of complex fractions. It is important to meticulously apply each step to achieve the correct derivative, as seen in the example where \( f(x) = \frac{x}{x+2} \).

Function Derivative

A function's derivative is essentially a way to compute its rate of change or slope at any given point, thus offering insight into the behavior of the function. In the case of a linear function, this derivative or rate of change remains constant. For more complex functions, such as the quotient of two functions, the approach is more nuanced.
When finding the derivative of a function, remember to:

• Identify the components of the function, as was done with \( u(x) = x \) and \( v(x) = x+2 \)
• Derive each component separately with respect to \( x \)

The derivative of the example function \( f(x) \), calculated using the Quotient Rule, results in \( f'(x) = \frac{2}{(x+2)^2} \). This tells us that for any value of \( x \), the slope of \( f(x) \) at that point will follow a specific pattern defined by this derivative.

Calculus

Calculus is a central branch of mathematics that focuses on rates of change and the accumulation of quantities. Central to its study are concepts like derivatives and integrals. Calculus allows us to go beyond static measures into dynamic analysis of variable change.
In the realm of calculus, derivatives are used to determine how a function changes as its input changes. This includes:

• Understanding motion and change, such as how the position of an object shifts over time
• Optimizing solutions in fields such as economics or engineering where variables fluctuate

In our specific exercise, calculus provides the tools needed to resolve derivatives through rules like the Quotient Rule. It equips learners with the skills to tackle rates of change in more natural, complex, and real-world situations. The power of calculus lies in its application, allowing us to solve diverse problems by grasping the continuous aspect of change.