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Magazine Chapter 1: Problem 57
Find \(\frac{d y}{d x}\) for each pair of functions.\( Find an equation for the tangent line to the graph of \)y=\sqrt{x^{2}+3 x}$ at the point (1,2).
Short Answer
Expert verified
The equation of the tangent line is \( y = \frac{5}{4}x - \frac{1}{4} \).
Step by step solution
01
1. Find the Derivative
First, find the derivative of the function using the chain rule. The function given is \(y = \sqrt{x^2 + 3x}\). Let \(u = x^2 + 3x\), then \(y = u^{1/2}\). Using the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). Now, \( \frac{dy}{du} = \frac{1}{2}u^{-1/2}\) and \( \frac{du}{dx} = 2x + 3\). Hence, \( \frac{dy}{dx} = \frac{1}{2}(x^2 + 3x)^{-1/2} \cdot (2x + 3)\).
02
2. Simplify the Derivative
Combine and simplify the terms to express the derivative in a more compact form: \( \frac{dy}{dx} = \frac{2x + 3}{2\sqrt{x^2 + 3x}} \).
03
3. Evaluate the Derivative at the Given Point
Substitute \(x = 1\) into the derivative to find the slope of the tangent line at the point (1,2). \( \frac{dy}{dx} \bigg|_{x = 1} = \frac{2(1) + 3}{2\sqrt{1^2 + 3(1)}} = \frac{5}{2\sqrt{4}} = \frac{5}{4} \). Thus, the slope of the tangent line at (1,2) is \( \frac{5}{4} \).
04
4. Determine the Tangent Line Equation
The equation of the tangent line can be found using the point-slope form of a line: \( y - y_1 = m(x - x_1) \), where \(m\) is the slope and \((x_1, y_1)\) is the given point. Plug in the values: \( y - 2 = \frac{5}{4}(x - 1) \). Simplify to get the equation in slope-intercept form: \( y = \frac{5}{4}x - \frac{1}{4} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a formula used to compute the derivative of a composite function. It's like breaking down a complex task into simpler steps. Suppose you have a function that is nested inside another function. To find its derivative, you'll separate them and handle them one by one.
For example, let's consider the function given: \(y = \sqrt{x^2 + 3x}\).
1. First, identify the inner function and the outer function. Here, the inner function is \(u = x^2 + 3x\), and the outer function is \(y = \sqrt{u}\).
2. Compute the derivative of the outer function with respect to the inner function: \( \frac{dy}{du} \).
3. Compute the derivative of the inner function with respect to \(x\): \( \frac{du}{dx} \).
4. Finally, use the chain rule formula which states: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
In our case: \( \frac{dy}{dx} = \frac{1}{2}(x^2 + 3x)^{-1/2} \cdot (2x + 3)\).
Chain rule simplifies complex differentiations, making it easier to handle nested functions.
For example, let's consider the function given: \(y = \sqrt{x^2 + 3x}\).
1. First, identify the inner function and the outer function. Here, the inner function is \(u = x^2 + 3x\), and the outer function is \(y = \sqrt{u}\).
2. Compute the derivative of the outer function with respect to the inner function: \( \frac{dy}{du} \).
3. Compute the derivative of the inner function with respect to \(x\): \( \frac{du}{dx} \).
4. Finally, use the chain rule formula which states: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
In our case: \( \frac{dy}{dx} = \frac{1}{2}(x^2 + 3x)^{-1/2} \cdot (2x + 3)\).
Chain rule simplifies complex differentiations, making it easier to handle nested functions.
Derivatives
Derivatives, in calculus, represent the rate at which a function is changing at any given point. Essentially, it is a measure of how a function's output value changes as its input changes.
For the function \(y = \sqrt{x^2 + 3x}\), finding the derivative involves breaking it into more manageable parts and then applying calculus rules, specifically the chain rule.
A derivative tells you the slope of the tangent line to the function at any point. For example, our derivative \( \frac{dy}{dx} = \frac{2x + 3}{2\sqrt{x^2 + 3x}} \) provides the slope for any value of \(x\).
This slope is particularly useful for solving various problems, such as finding the equation of a tangent line or understanding the behavior of the function at different points.
For the function \(y = \sqrt{x^2 + 3x}\), finding the derivative involves breaking it into more manageable parts and then applying calculus rules, specifically the chain rule.
A derivative tells you the slope of the tangent line to the function at any point. For example, our derivative \( \frac{dy}{dx} = \frac{2x + 3}{2\sqrt{x^2 + 3x}} \) provides the slope for any value of \(x\).
This slope is particularly useful for solving various problems, such as finding the equation of a tangent line or understanding the behavior of the function at different points.
Point-Slope Form
The point-slope form is a linear equation used to find the equation of a line when you're given a point on the line and the slope of the line.
The formula is: \( y - y_1 = m(x - x_1) \).
Here, \( (x_1, y_1) \) is the given point on the line and \(m\) is the slope.
In our solution, we found the slope at the point (1, 2) to be \( \frac{5}{4} \).
We then used this information in the point-slope form: \( y - 2 = \frac{5}{4}(x - 1) \).
This form is especially helpful because it's straightforward and gives a direct equation of a line when provided with a point and a slope.
The formula is: \( y - y_1 = m(x - x_1) \).
Here, \( (x_1, y_1) \) is the given point on the line and \(m\) is the slope.
In our solution, we found the slope at the point (1, 2) to be \( \frac{5}{4} \).
We then used this information in the point-slope form: \( y - 2 = \frac{5}{4}(x - 1) \).
This form is especially helpful because it's straightforward and gives a direct equation of a line when provided with a point and a slope.
Slope-Intercept Form
The slope-intercept form is another way to express the equation of a line. It's convenient because it directly shows the slope and y-intercept of the line.
The formula for slope-intercept form is: \( y = mx + b \).
Here, \( m \) is the slope of the line, and \( b \) is the y-intercept (the point where the line crosses the y-axis).
After using the point-slope form, we simplified the equation to the slope-intercept form: \( y = \frac{5}{4}x - \frac{1}{4} \).
This form makes it easy to identify the slope and intercept at a glance. You can instantly see that the slope is \( \frac{5}{4} \) and the y-intercept is \( -\frac{1}{4} \).
Using the slope-intercept form helps students and professionals graph linear equations quickly and understand certain properties of the line just by looking at the equation.
The formula for slope-intercept form is: \( y = mx + b \).
Here, \( m \) is the slope of the line, and \( b \) is the y-intercept (the point where the line crosses the y-axis).
After using the point-slope form, we simplified the equation to the slope-intercept form: \( y = \frac{5}{4}x - \frac{1}{4} \).
This form makes it easy to identify the slope and intercept at a glance. You can instantly see that the slope is \( \frac{5}{4} \) and the y-intercept is \( -\frac{1}{4} \).
Using the slope-intercept form helps students and professionals graph linear equations quickly and understand certain properties of the line just by looking at the equation.
