Question

In Exercises, find \(d y / d x\) implicitly. $$ x^{2} e^{-x}+2 y^{2}-x y=0 $$

Short answer

The derivative of the function, \(dy / dx\), is \(\frac{y - 2x e^{-x} + x^{2} e^{-x}}{4y - x}\).

Step-by-step solution

Differentiate the first term

Apply the product rule and the chain rule for differentiation to the first term, \(x^{2}e^{-x}\). The product rule states that the derivative of two functions multiplied together is the first function times the derivative of the second plus the second function times the derivative of the first. The derivative will be \((2x e^{-x} - x^{2} e^{-x})\).

Differentiate the second and third terms

The derivative of \(2y^{2}\) is \(4y dy/dx\), since y is a function of x. The derivative of \(-xy\) is \(- y - x dy/dx\), also applying the product rule.

Assemble the derivative

Combine all the differentiated elements to provide the derivative of the whole expression. So, \(2x e^{-x} - x^{2} e^{-x} + 4y(y') - y - x(y') = 0\).

Solve for \(dy/dx\)

The goal is to solve for \(dy/dx\), or \(y'\). Gather all the terms involving \(y'\) on one side of the equation and the remaining terms on the other side. Factor out \(y'\) and then divide to solve for \(y'\). This gives \(y' = \frac{y - 2x e^{-x} + x^{2} e^{-x}}{4y - x}\).

Key concepts

Product Rule

When finding derivatives of functions that are products of two or more other functions, the product rule is an essential tool. It states that the derivative of a product of two functions, say, \( f(x) \) and \( g(x) \), is given by \( f'(x)g(x) + f(x)g'(x) \). For example, to differentiate \( x^2 e^{-x} \), first identify \( f(x) = x^2 \) and \( g(x) = e^{-x} \). Applying the product rule, we find the derivative to be \( 2x e^{-x} - x^2 e^{-x} \) since the derivative of \( x^2 \) is \( 2x \) and the derivative of \( e^{-x} \) involves the chain rule, which will be discussed next.

In practical terms, remember to multiply the derivative of the first function by the second untouched function and add the product of the first untouched function and the derivative of the second function.

Chain Rule

The chain rule is a formula to calculate the derivative of a composite function. When a function \( y = g(f(x)) \) is the composition of two functions, the chain rule expresses the derivative \( dy/dx \) as \( g'(f(x)) \times f'(x) \). In the case of differentiating \( e^{-x} \), which is \( e \) raised to the power of \( -x \), we see a function inside another function—the exponential function with \( -x \) as its power.

To use the chain rule here, take the derivative of the outer function, which is \( e^u \) where \( u = -x \), yielding \( e^u \), and multiply it by the derivative of the inner function, which is \( -1 \). So, the chain rule gives us \( e^{-x} \times (-1) = -e^{-x} \), clearly illustrating how the composition of functions is handled in differentiation.

Derivative of Exponential Functions

Exponential functions have the general form \( f(x) = a^x \), where \( a \) is a constant. The derivative of an exponential function follows a unique pattern: \( \frac{d}{dx}a^x = a^x \ln(a) \), where \( \ln(a) \) is the natural logarithm of \( a \). However, when the base is the natural number \( e \), the derivative simplifies because \( \ln(e) = 1 \). Thus, for any function like \( e^{g(x)} \), employing the chain rule tells us that \( \frac{d}{dx} e^{g(x)} = e^{g(x)} \times g'(x) \).

This rule helps us quickly determine derivatives of natural exponential functions, an essential type of function in calculus used to describe growth and decay processes in real-world applications, like population growth, radioactive decay, and continuously compounded interest.

Solving Differential Equations

A differential equation is an equation that involves an unknown function and its derivatives. Solving a differential equation means finding the function that satisfies the equation. In the exercise provided, we employ implicit differentiation to find \( dy/dx \). Implicit differentiation treats \( y \) as an implicit function of \( x \) and allows us to differentiate both sides of the equation with respect to \( x \), even when it's not easy to solve for \( y \) explicitly.

After differentiating, we collect the terms involving \( dy/dx \) (notated as \( y' \)) on one side, facilitating the isolation of \( y' \) and ultimately solving for it. This process can handle a wide range of differential equations, beyond just straightforward separable or linear ones, making it an invaluable technique in fields such as engineering, physics, and economics where modeling change is crucial.