Problem 60
Question
Finding an Equation of a Tangent Line In Exercises \(55-62,\) find an equation of the tangent line to the graph of the function at the given point. $$ y=\ln \frac{e^{x}+e^{-x}}{2}, \quad(0,0) $$
Step-by-Step Solution
Verified Answer
The equation of the tangent line to the graph of the function at the point (2, 1) is \(y = \ln(5)x - 2\ln(5) + 1\).
1Step 1: Find the Derivative of the Given Function
Given function is \(y = 5^{x-2}\) The derivative of the function \(a^u\), where \(a\) is a constant and \(u\) is a function of \(x\), can be found using the formula \( \frac{d}{dx}a^u = \ln a * a^u * \frac{du}{dx}\). This lead us to \[\frac{dy}{dx}= \ln(5)*5^{x-2}\]
2Step 2: Find the Slope of the Tangent Line at Given Point
Substitute \(x=2\) into \(\frac{dy}{dx}\) to find the slope \(m\) of the tangent line at point (2,1). After substitution, we have \(m = \ln(5) *5^{2-2}=\ln(5)\).
3Step 3: Find the Equation of the Tangent Line
With the slope \(m\) and the given point (2,1), we can use the point-slope form equation which is \(y - y1 = m(x - x1)\), where \((x1, y1)\) represents a point on the line and \(m\) is the slope of the line. This gives us \(y - 1 = \ln(5) * (x - 2)\). After simplifying, we get the equation of the tangent line as \(y = \ln(5)x - 2\ln(5) + 1\).
Key Concepts
DerivativeExponential FunctionsPoint-Slope FormSlope of a Line
Derivative
The derivative is a fundamental concept in calculus. It measures how a function changes as its input changes. In simple terms, it tells us the slope of the function at any given point.
For a function that describes a curve, the derivative at a particular point can be thought of as the slope of the tangent line to the curve at that point. Derivatives are essential for finding rates of change and are widely used in physics, engineering, and beyond.
The process of finding a derivative is called differentiation. For example, in our exercise, we differentiated the function \(y = 5^{x-2}\) using a formula specific to exponential functions.
For a function that describes a curve, the derivative at a particular point can be thought of as the slope of the tangent line to the curve at that point. Derivatives are essential for finding rates of change and are widely used in physics, engineering, and beyond.
The process of finding a derivative is called differentiation. For example, in our exercise, we differentiated the function \(y = 5^{x-2}\) using a formula specific to exponential functions.
Exponential Functions
Exponential functions are mathematical functions of the form \(a^x\), where \(a\) is a constant and the exponent \(x\) is a variable. They are known for their unique characteristic of rapid growth.
Such functions frequently appear in real-world contexts including compound interest calculations, population growth models, and radioactive decay.
In our problem, the function \(y = 5^{x-2}\) is transformed using the property of exponents \(a^u\). The derivative was computed using the formula \(\frac{d}{dx}a^u = \ln a \cdot a^u \cdot \frac{du}{dx}\), which shows how exponential functions are handled in calculus.
Such functions frequently appear in real-world contexts including compound interest calculations, population growth models, and radioactive decay.
In our problem, the function \(y = 5^{x-2}\) is transformed using the property of exponents \(a^u\). The derivative was computed using the formula \(\frac{d}{dx}a^u = \ln a \cdot a^u \cdot \frac{du}{dx}\), which shows how exponential functions are handled in calculus.
Point-Slope Form
The point-slope form is a way to write the equation of a line. It's particularly useful when you know the slope of the line and a specific point the line passes through.
The formula is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope, and \((x_1, y_1)\) is the known point on the line. This form emphasizes how each coordinate on the line depends on its distance from a specific base point.
This method is pivotal in creating the equation of a tangent line, as seen in the exercise where \((2, 1)\) was used to plug values into the point-slope formula, resulting in the final tangent line equation.
The formula is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope, and \((x_1, y_1)\) is the known point on the line. This form emphasizes how each coordinate on the line depends on its distance from a specific base point.
This method is pivotal in creating the equation of a tangent line, as seen in the exercise where \((2, 1)\) was used to plug values into the point-slope formula, resulting in the final tangent line equation.
Slope of a Line
The slope of a line is a measure of its steepness and direction. It is calculated as the 'rise' over the 'run', which means how much the line goes up or down for a given horizontal distance.
The slope is a constant for a straight line, indicating a consistent rate of change. However, in calculus, the slope of a function at a point refers to the steepness of the tangent line at that point.
In the exercise, after computing the derivative, the slope at the specific point was found by substituting \(x=2\) into the derivative to get \(\ln(5)\). This slope was then used to determine the equation of the tangent line.
The slope is a constant for a straight line, indicating a consistent rate of change. However, in calculus, the slope of a function at a point refers to the steepness of the tangent line at that point.
In the exercise, after computing the derivative, the slope at the specific point was found by substituting \(x=2\) into the derivative to get \(\ln(5)\). This slope was then used to determine the equation of the tangent line.
Other exercises in this chapter
Problem 60
In Exercises 55–60, evaluate the integral. $$ \int_{0}^{\ln 2} 2 e^{-x} \cosh x d x $$
View solution Problem 60
Slope Field In Exercises \(57-60\) , use a computer algebra system to graph the slope field for the differential equation and graph the solution satisfying the
View solution Problem 60
In Exercises 41–64, find the derivative of the function. $$ f(x)=\ln \left(x+\sqrt{4+x^{2}}\right) $$
View solution Problem 60
Using Technology to Find an Integral In Exercises \(57-62\) use a computer algebra system to find or evaluate the integral. $$ \int \frac{x^{2}}{x-1} d x $$
View solution