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In Exercises 65 and 66 , use implicit differentiation to find an equation of the tangent line to the graph of the equation at the given point. $$1+\ln x y=e^{x-y}, \quad(1,1)$$

Short Answer

Expert verified
The equation of the tangent line to the graph of the equation at the given point \( (1, 1) \) is \( y = 1 \).

Step by step solution

01

Implicitly Differentiate the Equation

Start by implicitly differentiating the equation with respect to \(x\). The rule for differentiating a natural logarithm \( \ln(f(x)) \) is \(1/f(x) \cdot f'(x)\), and the rule for differentiating \(e^{g(x)}\) is \( e^{g(x)} \cdot g'(x)\).\n So the differentiated equation is: \[ \frac{1}{xy} \cdot (y + x \cdot y') = e^{x-y} \cdot (1 - y') \]
02

Solve for \(y'\)

Arrange the equation to solve for \( y' \) (the derivative of \( y \) with respect to \( x \)). This will involve combining all terms with \( y' \) on one side of the equation and the other terms on the opposite side. Doing so gives: \( y' = \frac{y - e^{x - y}}{x - e^{x - y}} \)
03

Substitute the Given Point \( (1, 1) \) to Find the Slope

Substitute the coordinates of the given point into \( y' \) to find the slope of the tangent line. This gives \( y'(1, 1) = \frac{1 - e^{1 - 1}}{1 - e^{1 - 1}} = 0 \). The slope of the tangent line is 0.
04

Use the Point-Slope Formula to Find the Equation of the Tangent Line

Use the point-slope form of a line given by \( y - y_1 = m(x - x_1) \), where \( m \) is the slope of the line and \( (x_1, y_1) \) is a point on the line. Substituting the slope and the point \( (1, 1) \) into the formula gives the equation of the tangent line as follows: \( y - 1 = 0(x - 1) \) or \( y = 1 \).

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Key Concepts

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

Tangent Line Equation
The tangent line to a curve at a given point is the straight line that just touches the curve at that point. It has the same slope as the curve at that point, which is given by the derivative of the function describing the curve. In simpler terms, imagine skimming a curve with a straight-edge ruler – the position where the ruler just grazes the curve without cutting through it represents the tangent line.

To find the equation of a tangent line, we need two key pieces of information: the slope of the curve at a point of tangency, which is calculated using the derivative, and the coordinates of the point of tangency. We then apply these values to the point-slope formula to derive the equation for the tangent line. This process involves calculus and algebra to pinpoint the exact slope and to express it in a y=mx+b or a similar linear format.
Differentiating Natural Logarithms
Differentiating natural logarithms involves understanding the fundamental behavior of the function \( \ln(x) \). The derivative of \( \ln(x) \) with respect to \( x \) is \( \frac{1}{x} \), based on the rule \( \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot u' \), where \( u \) is a function of \( x \) and \( u' \) is the derivative of \( u \).

This rule is crucial when dealing with functions involving \( \ln \) because it guides us through the process of finding the rate of change of logarithmic functions. The inherent property of the natural logarithm to be the inverse of the exponential function is what dictates this derivative relationship.
Differentiating Exponential Functions
On the other spectrum, we have differentiating exponential functions. These are functions where the variable appears in the exponent, such as \( e^x \). The derivative of \( e^x \) with respect to \( x \) is \( e^x \), which illustrates the unique property of the exponential function: it is its own derivative.

For more complex exponential functions of the form \( e^{g(x)} \), the rule \( \frac{d}{dx} e^{g(x)} = e^{g(x)} \cdot g'(x) \) is used. Here \( g(x) \) is a function of \( x \) and \( g'(x) \) is its derivative. This rule allows us to handle any compound exponential function by taking the original function and multiplying it by the derivative of the exponent.
Point-Slope Formula
The point-slope formula is a staple in coordinate geometry, particularly useful for writing the equation of a line when you know one point on the line and its slope. The formula is stated as \( y - y_1 = m(x - x_1) \), where \( m \) represents the slope, and \( (x_1, y_1) \) represents the coordinates of the known point.

This formula is derived from the slope definition and allows us to create an equation that represents all points on a line, not just the ones we already know. By plugging in the known values and simplifying, we get the desired linear equation, which is the basis for understanding linear relationships and solving a variety of problems in calculus and algebra.

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Most popular questions from this chapter

Logistic Differential Equation Show that solving the logistic differential equation \(\frac{d y}{d t}=\frac{8}{25} y\left(\frac{5}{4}-y\right), \quad y(0)=1\) results in the logistic growth function in Example \(7 .\) $$\left[Hint:\frac{1}{y\left(\frac{5}{4}-y\right)}=\frac{4}{5}\left(\frac{1}{y}+\frac{1}{\frac{5}{4}-y}\right)\right]$$

In Exercises 57-60, use a graphing utility to graph the slope field for the differential equation and graph the particular solution satisfying the specified initial condition. $$\begin{array}{l}{\frac{d y}{d x}=\frac{2 y}{\sqrt{16-x^{2}}}} \\\ {y(0)=2}\end{array}$$

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In Exercises 57-60, use a graphing utility to graph the slope field for the differential equation and graph the particular solution satisfying the specified initial condition. $$\begin{array}{l}{\frac{d y}{d x}=\frac{\sqrt{y}}{1+x^{2}}} \\\ {y(0)=4}\end{array}$$

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