Question

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$y=x e^{y}$$

Short answer

Question: Find the derivative of y with respect to x for the equation $$y = x e^y$$. Answer: $$ \frac{dy}{dx} = \frac{e^y}{1 - xe^y} $$

Step-by-step solution

Differentiate both sides implicitly with respect to x

Take the derivative of both sides of the equation with respect to x using implicit differentiation. $$\frac{d(y)}{d x}= \frac{d(x\cdot e^y)}{d x}$$.

Apply the product rule

Apply the product rule on the right side, where u(x) = x and v(x) = e^y and use the chain rule for v(x). The product rule states that the derivative of a product is: $$\frac{d(u \cdot v)}{d x} = u \frac{d v}{d x}+ \frac{d u}{d x} \cdot v$$ Now applying the product rule we get: $$\frac{d(y)}{d x}= x \cdot \frac{d(e^y)}{d x} + \frac{d(x)}{d x} \cdot e^y$$ .

Apply the chain rule

By applying the chain rule to the term \(\frac{d(e^y)}{d x}\), we get: $$ \frac{d(e^y)}{d x} = e^y \cdot \frac{d(y)}{d x} $$ since the derivative of \(e^y\) with respect to y is \(e^y\). Replace this term back into the equation: $$\frac{d(y)}{d x}= x \cdot(e^y \cdot \frac{d(y)}{d x}) + \frac{d(x)}{d x} \cdot e^y$$.

Simplify and solve for dy/dx

Simplify the equation and solve for dy/dx $$ \frac{d(y)}{d x} = x \cdot e^y \frac{d(y)}{d x} + e^y $$ Next, factor out the term dy/dx on the left side: $$ \frac{d(y)}{d x} (1 - xe^y) =e^y $$ Now, divide both sides by the expression (1 - xe^y) to obtain dy/dx: $$ \frac{d(y)}{d x} = \frac{e^y}{1 - xe^y}$$

Final Answer:

The derivative of y with respect to x is: $$ \frac{d y}{d x} = \frac{e^y}{1 - xe^y} $$

Key concepts

Chain Rule

The chain rule is a fundamental principle in calculus, used when taking the derivative of composite functions. Essentially, it allows us to differentiate a function that is nested within another function. Imagine peeling an onion, layer by layer, to see what elements comprise it—this is what the chain rule does for functions. The formal rule is: if you have a function \( g(f(x)) \), then the derivative of this composite function with respect to \( x \) is \( g'(f(x)) \cdot f'(x) \)—the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function itself. This is crucial when dealing with functions where one variable is dependent on another, such as in the exercise where the exponential function \( e^y \) depends on \( y \) which in turn depends on \( x \). Understanding this concept is pivotal for mastering implicit differentiation and advanced calculus problems.

Product Rule

When dealing with the multiplication of two functions, the product rule is the tool of choice for differentiation. It's an elegant shortcut avoiding the need to expand the entire function before differentiating. The rule states that for two functions \( u(x) \) and \( v(x) \), the derivative of their product \( u(x) \cdot v(x) \) is \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). You find the derivative of each function separately, and mash them together in a specific way: first \( u'(x) \), the derivative of the first, times the second function as is, plus the first function as is, times \( v'(x) \), the derivative of the second. It's a beautiful dance between the functions and their derivatives, and a classic move to use in problems such as our exercise, where the function \( x \cdot e^y \) involves both \( x \) and \( e^y \) multiplicatively.

Derivatives

Derivatives are the bread and butter of calculus, measuring how a function changes as its input changes. They're represented by the steepness or slope of the function at any given point, and are foundational for understanding change in a mathematical sense. To find the derivative of a function is to find this rate of change. In the context of our exercise, we are identifying how \( y \) changes in relation to \( x \), denoted as \( \frac{dy}{dx} \). Through the process of differentiation—applying rules like the product and chain rule—students can discover relationships between variables that aren't immediately obvious from the original equation. This concept underscores all of calculus and is essential for fields that require modeling change, from physics to economics.

Exponential Functions

Exponential functions, typified by \( e^x \)—where \( e \) is the mathematical constant approximately equal to 2.71828, are pivotal in mathematics due to their unique properties. They grow or decay at a rate proportional to their size, making them incredibly useful in modeling real-world phenomena like population growth or radioactive decay. The derivative of an exponential function is remarkable; \( e^x \) remains the same after differentiation: its own derivative is itself, which is beautifully simple yet profound. This self-derivative property also holds when the exponent is a function of \( x \), like \( e^y \), but in this case, multiplying by the derivative of the exponent, as seen when applying the chain rule. This concept not only appears in calculus courses but also plays a massive role in various scientific calculations and models.