Problem 46
Question
Find an equation for the tangent line to the graph at the specified value of \(x\) $$y=\left(x-\frac{1}{x}\right)^{3}, x=2$$
Step-by-Step Solution
Verified Answer
Tangent line equation is \(y = \frac{135}{16}x - \frac{108}{16}\).
1Step 1: Find the value of the function at the given x
Substitute the given value of \(x = 2\) into the function.\[y = \left(2 - \frac{1}{2}\right)^{3} = \left(\frac{4}{2} - \frac{1}{2}\right)^{3} = \left(\frac{3}{2}\right)^{3}\]\[y = \frac{27}{8}\]
2Step 2: Find the derivative of the function
To find the tangent line, we first need the derivative of the function. Let's compute \( y = \left( x - \frac{1}{x} \right)^{3} \) using the chain rule.Let \( u = x - \frac{1}{x} \). Then \( y = u^3 \).Differentiating \( u \) with respect to \( x \):\[\frac{du}{dx} = 1 + \frac{1}{x^2}\]Differentiating \( y \) with respect to \( u \):\[\frac{dy}{du} = 3u^2\]Using the Chain Rule:\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3u^2\left(1 + \frac{1}{x^2}\right)\]
3Step 3: Evaluate the derivative at x = 2
Now substitute \( x = 2 \) back into the expression for \( u \) to find \( u \) first:\[u = 2 - \frac{1}{2} = \frac{3}{2}\]Then substitute \( x = 2 \) and \( u = \frac{3}{2} \) into the derivative:\[\frac{dy}{dx}\bigg|_{x=2} = 3\left(\frac{3}{2}\right)^{2}\left(1 + \frac{1}{4}\right)\]\[= 3 \times \frac{9}{4} \times \frac{5}{4} = \frac{135}{16}\]
4Step 4: Write the equation of the tangent line
The equation of a tangent line is given by \( y - y_1 = m(x - x_1) \), where \(m\) is the slope, and \((x_1, y_1)\) is the point on the curve.From previous steps, we have:\(x_1 = 2\), \(y_1 = \frac{27}{8}\), and \(m = \frac{135}{16}\).Substituting these values:\[y - \frac{27}{8} = \frac{135}{16}(x - 2)\]This can be rearranged to:\[y = \frac{135}{16}x - \frac{135}{8} + \frac{27}{8}\]\[y = \frac{135}{16}x - \frac{108}{16}\]\[y = \frac{135}{16}x - \frac{108}{16}\]
Key Concepts
Derivative CalculationChain RuleFunction Evaluation
Derivative Calculation
Finding the derivative of a function is a core concept in calculus. Derivatives represent the rate of change of a function with respect to one of its variables. For instance, when you want the slope of a tangent line at a particular point on a curve, you use the derivative.
In our specific problem, we have the function \( y = \left( x - \frac{1}{x} \right)^3 \). To find its derivative, we need to identify how \( y \) changes as \( x \) changes. We start by substituting \( u = x - \frac{1}{x} \), making \( y = u^3 \).
Next, differentiate \( u \) concerning \( x \), which gives us \( \frac{du}{dx} = 1 + \frac{1}{x^2} \). Similarly, differentiate \( y \) concerning \( u \), resulting in \( \frac{dy}{du} = 3u^2 \). Combining these using the chain rule gives us the complete derivative \( \frac{dy}{dx} = 3u^2 \left( 1 + \frac{1}{x^2} \right) \).
Understanding derivatives will help you describe the slope of the line that just touches a curve at a point -- the tangent line.
In our specific problem, we have the function \( y = \left( x - \frac{1}{x} \right)^3 \). To find its derivative, we need to identify how \( y \) changes as \( x \) changes. We start by substituting \( u = x - \frac{1}{x} \), making \( y = u^3 \).
Next, differentiate \( u \) concerning \( x \), which gives us \( \frac{du}{dx} = 1 + \frac{1}{x^2} \). Similarly, differentiate \( y \) concerning \( u \), resulting in \( \frac{dy}{du} = 3u^2 \). Combining these using the chain rule gives us the complete derivative \( \frac{dy}{dx} = 3u^2 \left( 1 + \frac{1}{x^2} \right) \).
Understanding derivatives will help you describe the slope of the line that just touches a curve at a point -- the tangent line.
Chain Rule
The chain rule is a fundamental tool in calculus used to compute the derivative of composite functions. A composite function is one formed by "substituting" one function within another.
For example, if you have \( y = (u)^3 \) where \( u = x - \frac{1}{x} \), then \( y \) is a function of \( u \), and \( u \) is a function of \( x \). The chain rule allows us to compute \( \frac{dy}{dx} \) by multiplying the derivative of \( y \) with respect to \( u \) by the derivative of \( u \) with respect to \( x \).
This gives us the complete derivative: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3u^2 \left( 1 + \frac{1}{x^2} \right) \]
Without the chain rule, differentiating composite functions would be a cumbersome task. Breaking functions into smaller, manageable parts eases the differentiation process significantly.
For example, if you have \( y = (u)^3 \) where \( u = x - \frac{1}{x} \), then \( y \) is a function of \( u \), and \( u \) is a function of \( x \). The chain rule allows us to compute \( \frac{dy}{dx} \) by multiplying the derivative of \( y \) with respect to \( u \) by the derivative of \( u \) with respect to \( x \).
This gives us the complete derivative: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3u^2 \left( 1 + \frac{1}{x^2} \right) \]
Without the chain rule, differentiating composite functions would be a cumbersome task. Breaking functions into smaller, manageable parts eases the differentiation process significantly.
Function Evaluation
Function evaluation is the process of determining the output of a function for a particular input. This concept helps you understand how a function behaves at specific points, which is crucial in calculus, especially when finding tangent lines to curves.
In this exercise, we started with the function \( y = \left( x - \frac{1}{x} \right)^3 \) and a given \( x = 2 \). Evaluating the function at \( x = 2 \) involves substituting this value into the function.
Substituting gives: \[ y = \left( 2 - \frac{1}{2} \right)^3 = \left( \frac{3}{2} \right)^3 \]
Calculating further yields \( y = \frac{27}{8} \). This value represents the point on the curve at \( x = 2 \), which is essential for finding the tangent line. Evaluating functions is a straightforward yet pivotal part of solving calculus problems, especially in practical applications such as calculating slopes or rates of change at a given point.
In this exercise, we started with the function \( y = \left( x - \frac{1}{x} \right)^3 \) and a given \( x = 2 \). Evaluating the function at \( x = 2 \) involves substituting this value into the function.
Substituting gives: \[ y = \left( 2 - \frac{1}{2} \right)^3 = \left( \frac{3}{2} \right)^3 \]
Calculating further yields \( y = \frac{27}{8} \). This value represents the point on the curve at \( x = 2 \), which is essential for finding the tangent line. Evaluating functions is a straightforward yet pivotal part of solving calculus problems, especially in practical applications such as calculating slopes or rates of change at a given point.
Other exercises in this chapter
Problem 45
Find an equation for the tangent line to the graph at the specified value of \(x\) $$y=\sec ^{3}\left(\frac{\pi}{2}-x\right), x=-\frac{\pi}{2}$$
View solution Problem 45
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View solution Problem 46
Show that \(f(x)\) is continuous but not differentiable at the indicated point. Sketch the graph of \(f\) (a) \(f(x)=\sqrt[3]{x}, x=0\) (b) \(f(x)=\sqrt[3]{(x-2
View solution Problem 46
Find (a) \(y^{\prime \prime \prime}(0),\) where \(y=4 x^{4}+2 x^{3}+3\) (b) \(\left.\frac{d^{4} y}{d x^{4}}\right|_{x=1},\) where \(y=\frac{6}{x^{4}}\)
View solution