Problem 64 - University Calculus: Early Transcendentals

Question

Find \(d y / d x\). $$\ln x y=e^{x+y}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{xy e^{x+y} - y}{x - xy e^{x+y}} \)

Step 1: Differentiate both sides with respect to x

The equation given is \( \ln(xy) = e^{x+y} \). We need to differentiate both sides of the equation with respect to \( x \). For the left-hand side, we use the chain rule and the product rule, knowing that the derivative of \( \ln u \) is \( \frac{1}{u} \frac{du}{dx} \). For the right-hand side, we also use the chain rule for \( e^{u} \) where \( \frac{d}{dx}(e^u) = e^u \frac{du}{dx} \).

Step 2: Apply the chain rule to the left side

First, differentiate \( \ln(xy) \) with respect to \( u = xy \), which gives \( \frac{1}{xy} \cdot \frac{d}{dx}(xy) \). Then, use the product rule on \( xy \) to obtain \( x \frac{dy}{dx} + y \). This gives the expression \( \frac{1}{xy}(x \frac{dy}{dx} + y) \).

Step 3: Differentiate the right side using the chain rule

Using the chain rule, differentiate \( e^{x+y} \). This gives \( e^{x+y} \cdot (1 + \frac{dy}{dx}) \) since the derivative of \( x+y \) with respect to \( x \) is \( 1+\frac{dy}{dx} \).

Step 4: Set the derivatives equal

From Steps 2 and 3, set the differentiated expressions equal to each other: \( \frac{1}{xy}(x \frac{dy}{dx} + y) = e^{x+y}(1 + \frac{dy}{dx}) \).

Step 5: Simplify and solve for \( \frac{dy}{dx} \)

Multiply both sides of the equation by \( xy \) to clear the fraction: \( x \frac{dy}{dx} + y = xy e^{x+y}(1 + \frac{dy}{dx}) \). Expand the right side: \( xy e^{x+y} + xy e^{x+y} \frac{dy}{dx} \). Collect terms involving \( \frac{dy}{dx} \): \( x \frac{dy}{dx} - xy e^{x+y} \frac{dy}{dx} = xy e^{x+y} - y \). Factor \( \frac{dy}{dx} \) out: \( \frac{dy}{dx}(x - xy e^{x+y}) = xy e^{x+y} - y \). Solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{xy e^{x+y} - y}{x - xy e^{x+y}} \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. It's like peeling an onion; each layer has to be dealt with step by step. If you have a function inside another function, use the chain rule to find the derivative.
To apply it, identify the "outer" and "inner" functions. For example, in a function like \( \ln(xy) \), \ the outer function is \ \(\ln(u)\) and \( u = xy \) is the inner one.
- Begin by differentiating the outer function, leaving the inner one intact.- Multiply this by the derivative of the inner function.
In our exercise, the chain rule is used on both sides of the equation \( \ln(xy)=e^{x+y} \).
On the left, \ \(\ln(u)\) becomes \ \(\frac{1}{u}\) when differentiated, but remember to multiply it by the derivative of \(u = xy \).
On the right, \ \(e^{u}\) maintains its form \ and is multiplied by the derivative of the exponent \( x+y \). This careful application allows us to manage complex combinations of functions simply and effectively.

Product Rule

The product rule is another key concept in differentiation, used when differentiating a product of two functions. Think of it as splitting a job between you and a partner where you handle the change in one part while your partner handles the other.
The formula for the product rule is straightforward: \( (fg)' = f'g + fg' \), where \( f \) and \( g \) are functions of \( x \).
When applying it to problems such as in our given exercise, it helps deal with parts like \( xy \) on the left.
- Take the first function, \(x\), and differentiate it, multiplying by the second function \(y\).- Then, do the opposite operation: leave the first function \(x\) as is and differentiate \(y\).
In this way, using the product rule correctly ensures that all parts of a multiplication relationship are accounted for, providing a clear path to finding the correct derivative.

Implicit Differentiation

Implicit differentiation is a useful technique when dealing with equations where \( y \) is not isolated. Instead of solving for \( y \) explicitly before differentiating, we differentiate both sides of the equation with respect to \( x \) at once.
This method is particularly handy for relationships like \( \ln(xy) = e^{x+y} \), where isolating \( y \) could be cumbersome.
- Start by differentiating all terms in the equation. Treat \( y \) as a function of \( x \) and remember to apply the chain rule.- Every time you differentiate something with \( y \), tack on \( \frac{dy}{dx} \) because you're differentiating \( y \) with respect to \( x \).
So, when we differentiate terms like \( \ln(xy) \) and \( e^{x+y} \), we retain \( \frac{dy}{dx} \) whenever \( y \) appears.
This preserves the dependency between \( y \) and \( x \) and allows us to find \( \frac{dy}{dx} \) even in complex equations, leading us to a solution that highlights how changes in \( x \) affect \( y \). This step ensures that we're correctly capturing the relationship between variables as they change.