Problem 49 - Calculus Single Variable

Question

Find the tangent line to the graph of the given function at the given point. $$ f(x)=\sin (x) /(x+1), \quad P=(0,0) $$

Step-by-Step Solution

Answer

The tangent line at \(P\) is \(y = x\).

Step 1: Find the Function's Value at Point P

We begin by ensuring that the point \(P = (0, 0)\) lies on the function \(f(x) = \frac{\sin(x)}{x+1}\). By evaluating \(f(0)\), we find \(f(0) = \frac{\sin(0)}{0+1} = \frac{0}{1} = 0\). Hence, \(P\) is on the graph of \(f\).

Step 2: Calculate the Derivative of the Function

Next, we need to find the derivative of the function \(f(x)\) to determine the slope of the tangent line. Using the quotient rule, if \(u(x) = \sin(x)\) and \(v(x) = x+1\), then \(f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}\). Thus, \(u'(x) = \cos(x)\) and \(v'(x) = 1\), leading to:\[ f'(x) = \frac{(x+1)\cos(x) - \sin(x)}{(x+1)^2} \]

Step 3: Evaluate the Derivative at Point P

We evaluate the derivative \(f'(x)\) at \(x = 0\) to find the slope of the tangent line at point \(P\). Substituting \(x = 0\) into the derivative:\[ f'(0) = \frac{(0+1)\cos(0) - \sin(0)}{(0+1)^2} = \frac{1 \cdot 1 - 0}{1^2} = 1 \]The slope of the tangent line at \(P\) is 1.

Step 4: Write the Equation of the Tangent Line

Using the point-slope form \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point \(P = (0,0)\) and \(m = 1\), the equation becomes:\[ y - 0 = 1(x - 0) \] Simplifying gives:\[ y = x \]This is the equation of the tangent line.

Key Concepts

Derivative Calculation

Understanding how to calculate the derivative of a function is crucial in finding the slope of a tangent line. The derivative essentially measures how a function changes as its input changes. For the given function, \( f(x) = \frac{\sin(x)}{x+1} \), we need to find the derivative \( f'(x) \). This requires us to differentiate both the numerator and the denominator separately using specific rules of differentiation. Calculating a derivative allows us to determine how steep or flat a graph is at any given point, and it's the foundation for other calculations like finding a tangent line. Once you have the derivative, you can evaluate it at a particular point to find the slope of the tangent line.

Quotient Rule

When differentiating a function that is the quotient of two other functions, we use the quotient rule. This is particularly helpful for rational functions like \( \frac{\sin(x)}{x+1} \). According to the quotient rule, the derivative \( f'(x) \) is found by:

• Taking the derivative of the numerator \( \sin(x) \) which is \( \cos(x) \).
• Taking the derivative of the denominator \( x+1 \) which is 1.
• Applying the formula: \( f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \).

This rule is essential in calculus because it provides a method to deal with complicated fractions in a systematic way, allowing us to find their derivatives accurately. It's important to remember each step, as missing one can lead to incorrect results.

Slope of Tangent Line

The slope of a tangent line is found by evaluating the derivative at a specific point. In our example, the point \( P = (0,0) \) is used. By substituting \( x = 0 \) into the derivative \( f'(x) = \frac{(x+1)\cos(x) - \sin(x)}{(x+1)^2} \), we find the slope at \( P \) to be 1. The slope tells us the rate of change at that exact point on the graph. A slope of 1 means that for every unit the x-value increases, the y-value increases by the same amount. This is a direct way to visually and numerically comprehend the behavior and direction of a function at a single point.

Point-Slope Form

The point-slope form of a linear equation is a handy formula, especially when you know one point on a line and its slope. It is written as \( y - y_1 = m(x - x_1) \). In our scenario, the point \( P = (0,0) \) and the slope \( m = 1 \) make the use of this form straightforward. Substituting these into the formula gives us: \[ y - 0 = 1(x - 0) \] Simplifying, we get \( y = x \), which is the equation of the tangent line at point \( P \). The beauty of point-slope form lies in its simplicity – once you have the point and the slope, writing the equation is just a matter of substitution.