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Chapter 7: Problem 83

Find an equation of the tangent line to the graph of the equation \(y=x e^{1 x^{3}}+\ln \left|2-x^{2}\right|\) at the point \(P(1, e) .\)

Short Answer

Expert verified
The equation of the tangent line is \( y = (-2e-2)x + 3e + 2 \).

Step by step solution

01

Find the Derivative

To find the equation of the tangent line, we first need to determine the derivative of the function at the given point. The function is \( y = x e^{1/x^3} + \ln |2 - x^2| \). We will find \( y' \) using derivative rules for exponential and logarithmic functions.
02

Differentiate the First Term

For the first term \( x e^{1/x^3}\), use the product rule: if \( u = x \) and \( v = e^{1/x^3} \), the derivative \( \frac{d}{dx}(uv) = u'v + uv' \). Compute each part: \( u' = 1 \) and \( v' = e^{1/x^3}\left(-\frac{3}{x^4}\right) \). So, the derivative is: \( 1 \cdot e^{1/x^3} + x \cdot \left(-\frac{3}{x^4}\right)e^{1/x^3} \).
03

Differentiate the Second Term

For the second term \( \ln |2-x^2| \), use the chain rule together with the rule for logarithms: the derivative is \( \frac{1}{2-x^2}\cdot (-2x) \), which simplifies to \( \frac{-2x}{2-x^2}\).
04

Combine Derivatives

Combine the derivatives from the previous two steps to get the complete derivative of the function. The complete derivative is: \( y' = e^{1/x^3} - \frac{3}{x^3} e^{1/x^3} - \frac{2x}{2-x^2} \). Simplify as needed.
05

Evaluate the Derivative at the Point

Substitute \( x = 1 \) into the derivative \( y' \) to find the slope of the tangent line at the point \( P(1, e) \). Calculate: \( e^{1} - \frac{3}{1}e^{1} - \frac{2 \cdot 1}{2-1^2} = e - 3e - 2 = -2e - 2 \).
06

Use Point-Slope Form to Find the Tangent Line

Use the point-slope form of the line equation, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope calculated in Step 5, \( x_1 = 1 \), and \( y_1 = e \). Substitute these values: \( y - e = (-2e-2)(x - 1) \).
07

Simplify the Tangent Line Equation

Expand and simplify the equation from Step 6 to get the final equation of the tangent line. The simplified form is \( y = (-2e-2)x + (2e + 2 + e) \). Therefore, \( y = (-2e-2)x + 3e + 2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
A derivative is a core concept in calculus that measures how a function changes as its input changes. In simpler terms, it represents the slope of the tangent line to a curve at a given point. Calculating a derivative helps us understand rates of change, such as velocity in physics or growth rates in biology.
  • The derivative of a function is denoted as \( y' \) or \( \frac{dy}{dx} \).
  • For a function \( y = f(x) \), the derivative at a point tells us the slope of the function at that particular x-value.
To find the equation of a tangent line, like in our exercise, we start by differentiating the function. This involves using various rules such as the product rule and chain rule, depending on the complexity of the function.
Product Rule
The product rule is essential when differentiating functions that are products of two or more functions. If we have a function \( y = u(x) \cdot v(x) \), the product rule states that its derivative is \( u'v + uv' \). This means we differentiate each function separately and then sum the products.
  • Let \( u(x) \) and \( v(x) \) be differentiable functions.
  • The derivative of their product is \( u'v + uv' \).
In our exercise, we applied the product rule to the term \( x e^{1/x^3} \). The derivative of \( x \) is 1, and the derivative of \( e^{1/x^3} \) is found using the chain rule. Combining these results gives us the derivative of the product.
Chain Rule
The chain rule is crucial when dealing with composite functions, where one function is inside another. For a composite function \( y = f(g(x)) \), the chain rule helps find the derivative by multiplying the derivative of the outer function by the derivative of the inner function. It is written as \( f'(g(x)) \cdot g'(x) \).
  • Identify the outer function \( f \) and the inner function \( g \).
  • Calculate \( f'(g(x)) \) and \( g'(x) \).
In our specific problem, the chain rule was employed to differentiate \( e^{1/x^3} \). Here, \( f(u) = e^u \) where \( u = 1/x^3 \). The derivative becomes \( e^{1/x^3} \cdot \left(-\frac{3}{x^4}\right) \). This is then used as part of the product rule calculation.
Point-Slope Form
The point-slope form is a common way to write the equation of a line. It is especially useful when you have one point on the line and the slope. The standard format is \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a point on the line and \( m \) is the slope.
  • Identify the slope \( m \) at the desired point.
  • Use the coordinates of the known point \( (x_1, y_1) \).
  • Plug in these values to form the equation.
In our exercise, once we calculated the slope \( m = -2e - 2 \), we substituted it along with the point \( (1, e) \) into the point-slope form. Then, by simplifying, we arrived at the tangent line equation \( y = (-2e-2)x + 3e + 2 \). This equation represents the tangent line to the curve at the given point.

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