Problem 237 - Calculus 3

Question

For the following exerises, find \(\frac{d y}{d x}\) using parial derivatives. $$ e^{x y}+y e^{y}=1 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the function structure

The given expression is \(e^{xy} + ye^y = 1\). We need to find \(\frac{dy}{dx}\) using partial derivatives. This expression implicitly defines \(y\) as a function of \(x\).

Step 2: Apply implicit differentiation

Differentiate both sides of the equation with respect to \(x\). For the term \(e^{xy}\), use the chain rule: \(\frac{d}{dx}(e^{xy}) = e^{xy}(y + x\frac{dy}{dx})\). For the term \(ye^y\), use the product rule and chain rule: \(\frac{d}{dx}(ye^y) = e^y\frac{dy}{dx} + y e^y \frac{dy}{dx}\).

Step 3: Differentiate the right-hand side

The right-hand side of the equation is \(1\), which is constant. Its derivative with respect to \(x\) is \(0\).

Step 4: Substitute derivatives into the equation

Substituting the derivatives into the original equation, we get: \[ e^{xy}(y + x\frac{dy}{dx}) + e^y\frac{dy}{dx} + y e^y \frac{dy}{dx} = 0. \]

Step 5: Collect terms involving \(\frac{dy}{dx}\)

Rearrange terms to isolate \(\frac{dy}{dx}\): \[ e^{xy} y + (xe^{xy} + e^y + ye^y)\frac{dy}{dx} = 0. \]

Step 6: Solve for \(\frac{dy}{dx}\)

Isolate \(\frac{dy}{dx}\) to find the derivative: \[ (xe^{xy} + e^y + ye^y)\frac{dy}{dx} = -e^{xy} y. \]Finally, divide both sides by \(xe^{xy} + e^y + ye^y\) to get:\[ \frac{dy}{dx} = \frac{-e^{xy} y}{xe^{xy} + e^y + ye^y}. \]

Key Concepts

Partial Derivatives

Partial derivatives are an extension of regular derivatives to functions of multiple variables. They allow us to take the derivative of a function concerning one variable while keeping other variables constant. This technique is particularly useful when dealing with equations that implicitly define one variable in terms of others, as is often seen in implicit differentiation.

In the example given, the expression is treated as a function involving both variables, where we casually regard it's response in various separate axes:

• For instance, when differentiating with respect to \(x\), we treat \(y\) as a constant, and vice-versa.
• This makes it feasible to compute the change as the input in one direction varies, thereby unraveling connections between variables in entangled equations.

Chain Rule

The chain rule is a fundamental tool in calculus used for finding the derivative of composite functions, where a function is nested within another. It is indispensable when dealing with implicit differentiation, allowing us to differentiate effectively step-by-step as functions are linked.

Here’s how the chain rule operates:

• The inner function is differentiated first.
• Then, the outer function is differentiated, and both are multiplied to yield the result.

In the equation \(e^{xy}\), we differentiate the inside (\(xy\)) using the product rule, and use chain rule for the outer function, yielding \(e^{xy}(y + x\frac{dy}{dx})\). This engagement of both rules within the chain process reveals intricate derivations of nested functions, empowering us to trace through multiple layers in implicit differentiation.

Product Rule

The product rule is a technique in calculus for finding the derivative of the product of two functions. In situations where two parts of a function depend on a variable, the result isn't as simple as taking individual derivatives separately.

For example, in the term \(ye^y\), both \(y\) and \(e^y\) must be differentiated. The product rule dictates that:

• The derivative of the first function times the second function remains unaltered, plus
• The first function untouched times the derivative of the second function

So, \(\frac{d}{dx}(ye^y) = e^y\frac{dy}{dx} + ye^y\frac{dy}{dx}\). This highlights how intertwining variables compel careful dissection, leveraging product rule in tandem with chain rule to handle intertwined terms comprehensively.

Function of a Function

In calculus, when a function is composed within another function, it's termed as a function of a function - a concept pivotal for implicit differentiation and finding derivatives in layered scenarios.

Consider \(e^{xy}\): It’s a prime example where \(e\) is the outer function, and \(xy\) is the inner function. The inner function \(xy\) itself stems from the multiplication of \(x\) and \(y\), aspectually addressed via chain and product rules.

Identifying nested layers and unpacking them as simpler constituents is key to simplifying the derivative process:

• Dive into the structure till independent manageable layers are visible.
• Analyze connections and use rule applications like chain and product rules.

Untwining these intricate relationships allows us to navigate involving entwined or entangled derivatives, a hallmark in proficiency with implicit differentiation.