Problem 24 - Calculus Single Variable

Question

Use implicit differentiation to find the tangent line to the given curve at the given point \(P_{0}\). \(x e^{x-1}-\ln (x y)=1 \quad P_{0}=(1,1)\)

Step-by-Step Solution

Answer

The equation of the tangent line at \((1,1)\) is \(y = x\).

Step 1: Differentiate Implicitly

To differentiate implicitly, let's differentiate both sides of the equation: \(x e^{x-1} - \ln(x y) = 1\). First, differentiate \(x e^{x-1}\):\(\frac{d}{dx}(x e^{x-1}) = e^{x-1} + x e^{x-1}\). Next, differentiate \(\ln(x y)\) using the chain rule: \(\frac{d}{dx}(\ln(x y)) = \frac{1}{xy} \cdot (y + x \frac{dy}{dx})\). Putting everything together, the differentiated equation is:\[e^{x-1} + x e^{x-1} - \frac{1}{xy}(y + x \frac{dy}{dx}) = 0\].

Step 2: Solve for \(\frac{dy}{dx}\)

Now, rearrange the equation to solve for \(\frac{dy}{dx}\):\[e^{x-1} + x e^{x-1} = \frac{y + x \frac{dy}{dx}}{xy}\].Multiplying through by \(xy\) gives:\[xy(e^{x-1} + x e^{x-1}) = y + x \frac{dy}{dx}\].Isolate \(x \frac{dy}{dx}\):\[xy(e^{x-1} + x e^{x-1}) - y = x \frac{dy}{dx}\].Finally, solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{xy(e^{x-1} + x e^{x-1}) - y}{x}\].

Step 3: Evaluate \(\frac{dy}{dx}\) at \((1,1)\)

Substitute \(x=1\) and \(y=1\) into the equation for \(\frac{dy}{dx}\):\[\frac{dy}{dx}\bigg|_{(1,1)} = \frac{1 \cdot 1 (e^{1-1} + 1 \cdot e^{1-1}) - 1}{1}\].Simplify the expression:\[= \frac{1(1 + 1) - 1}{1} = \frac{2 - 1}{1} = 1\].Thus, the slope of the tangent line at the point \((1,1)\) is 1.

Step 4: Find the Equation of the Tangent Line

Using the point-slope form of the equation of a line, we have:\[y - y_1 = m(x - x_1)\].Here, \((x_1, y_1) = (1, 1)\) and the slope \(m = 1\):\[y - 1 = 1(x - 1)\].Simplifying gives:\[y = x\].Thus, the equation of the tangent line at \((1,1)\) is \(y = x\).

Key Concepts

Tangent Line

A tangent line to a curve at a particular point provides an immediate view of the curve's behavior at that point. Essentially, it's the straight line that just "touches" the curve without cutting through it. It gives us a linear approximation of the curve near that point. For any given function or curve, the equation of the tangent line can be determined by identifying the slope of the line and a point through which the line passes.
A tangent line is often described by:

• Point: This is the specific point on the curve where the tangent line is to be determined. In our example, it is the point \((1, 1)\).
• Slope: This is the steepness or inclination of the tangent line. It’s calculated using the derivative of the function. The derivative essentially measures how the function is changing at a specific point.
• Equation: We use the slope \(m\) and the point to write the equation of the tangent line, often in the point-slope form: \(y - y_1 = m(x - x_1)\).

In the exercise, by using implicit differentiation, we found the slope of the tangent line at the point \((1, 1)\) to be 1. Thus, the equation of the tangent line at this point was calculated to be \(y = x\).

Chain Rule

The chain rule is a fundamental technique in calculus for differentiating composite functions. It's particularly useful when dealing with functions of functions, or when variables are not easily separated, which often happens in implicit differentiation.
When you differentiate implicitly, as we did with the equation \(x e^{x-1} - \ln (xy) = 1\), you frequently encounter terms where one variable is expressed in terms of another—like \(\ln(xy)\). The derivative of \(\ln(xy)\) requires the chain rule because it’s essentially a function within a function.

• Basic Concept: If you have a function \(g(u(x))\), the derivative using the chain rule is \(g'(u(x)) \cdot u'(x)\).
• Application to \(\ln(xy)\): Here, \(\ln(xy)\) is a composite function; therefore, to differentiate it, we find the derivative of the outer function \(\ln\) (which is \(1/z\) where \(z = xy\)), and multiply it by the derivative of the inner function \(xy\), resulting in \(\frac{1}{xy} \cdot (y + x \frac{dy}{dx})\).

Mastering the chain rule is essential for working through problems involving implicit differentiation, as it enables you to navigate complicated expressions smoothly.

Slope of the Tangent

The slope of the tangent line is a central aspect of determining the tangent line itself. It essentially tells us how steep the line is at a given point on the curve. Utilizing implicit differentiation helps us calculate this value, especially when the relationship between variables isn't straightforward.
Here’s a simple breakdown of how to find the slope of a tangent for the presented exercise:

• Implicit Differentiation: Begin by differentiating both sides of the given equation \(x e^{x-1} - \ln(xy) = 1\) with respect to \(x\). This process, implicitly, accounts for \(y\) as a function of \(x\).
• Solve for \(\frac{dy}{dx}\): After differentiating, rearrange the equation to isolate and solve for \(\frac{dy}{dx}\). This derivative represents the rate of change of \(y\) with respect to \(x\) and thus gives the slope of the tangent line at the point.
• Evaluate at Specific Point: Substitute your point \((1, 1)\) into the derivative \(\frac{dy}{dx}\) to find the exact numeric value of the slope at that point. For our curve, this turned out to be 1, suggesting the tangent rises 1 unit vertically for each 1 unit it moves horizontally.

Understanding how to find the slope is indispensable for constructing the equation of a tangent line and interpreting the behavior of curves.