Problem 43 - Calculus

Question

If \(d y / d x=y\) find the derivative of \(e^{-x} y\) by the product rule. Deduce that \(y(x)=C e^{x}\) for some constant \(C\).

Step-by-Step Solution

Answer

The derivative of \(e^{-x} y\) is 0, implying \(y(x) = C e^{x}\) for some constant \(C\).

Step 1: Product Rule Introduction

To find the derivative of the product of two functions, we'll use the product rule. The product rule states that if we have two functions, say, \(u(x)\) and \(v(x)\), then the derivative of their product is: \((u(x)v(x))' = u'(x)v(x) + u(x)v'(x)\). In this problem, we treat \(u(x) = e^{-x}\) and \(v(x) = y\).

Step 2: Differentiate \(u(x) = e^{-x}\)

First, we need to differentiate \(u(x) = e^{-x}\). Using the chain rule, the derivative is \(u'(x) = -e^{-x}\).

Step 3: Differentiate \(v(x) = y\) Using Given Equation

We know from the problem statement that \(\frac{dy}{dx} = y\). Thus, the derivative of \(v(x) = y\) is simply \(v'(x) = y\).

Step 4: Apply Product Rule

Now that we have \(u'(x) = -e^{-x}\) and \(v'(x) = y\), we can apply the product rule: \(\frac{d}{dx}(e^{-x} y) = u'(x)v(x) + u(x)v'(x) = (-e^{-x})y + e^{-x}y\).

Step 5: Simplify

Simplify the expression for the derivative: \(-e^{-x} y + e^{-x} y = 0\). This means that the derivative of \(e^{-x} y\) is equal to zero.

Step 6: Deduce the Form of \(y(x)\)

Since the derivative \(\frac{d}{dx}(e^{-x} y) = 0\), it implies that \(e^{-x} y\) is a constant. Let's denote this constant by \(C\). Therefore, \(e^{-x} y = C\).

Step 7: Solve for \(y(x)\)

To find \(y(x)\), solve \(e^{-x} y = C\) for \(y\): multiply both sides by \(e^{x}\), giving \(y = C e^{x}\). This is the general solution for \(y(x)\), where \(C\) is a constant.

Key Concepts

Product Rule

When dealing with the derivative of a product of functions, the product rule plays a vital role. It helps us find derivatives easily. The product rule states: if you have two functions, say \(u(x)\) and \(v(x)\), the derivative of their product is given by \((u(x)v(x))' = u'(x)v(x) + u(x)v'(x)\).

Here's how to apply it efficiently:

• Identify the two functions involved in the product.
• Find the derivative of each function separately.
• Combine these derivatives using the product rule formula.

In our example exercise, the functions \(u(x)\) and \(v(x)\) are chosen as \(e^{-x}\) and \(y\), respectively. Each derivative is then found and plugged into the product rule formula. This step makes differentiating complex multiplied functions straightforward.

Derivative

Derivatives show how a function changes at any given point. They are the backbone of differentiation and integral to calculus. Essentially, a derivative tells us the rate of change or the slope of the function at a point.

Finding derivatives involves rules and techniques such as the product rule, chain rule, and more. In our example, we need to know specific derivatives:

• Derivative of exponential functions.
• How to differentiate a variable using differential equations.

For the function \(u(x) = e^{-x}\), we calculate its derivative using the negative exponent. The derivative is \(-e^{-x}\). Similarly, for \(v(x) = y\), it already provides \(\frac{dy}{dx} = y\), since it is stated in the problem.

Recognizing how these derivatives interact is key, especially when applying rules.

Chain Rule

The chain rule is used when differentiating composite functions. It breaks down the process into more manageable steps, which can simplify finding derivatives of functions that are composed of other functions.

To use the chain rule effectively:

• Identify the inner function and the outer function.
• Differentiate both functions separately.
• Multiply the derivative of the outer function by the derivative of the inner function.

In our exercise, the chain rule is required to find the derivative of the exponential function \(e^{-x}\). Here, \( e^{-x}\) is treated as a composition of \(e^x\) and \(-x\). By differentiating \( -x\) to get \(-1\) and applying the rule to \(e^x\), we successfully find \(-e^{-x}\).

This simplification allows us to apply our chain rule knowledge directly to complex problems involving differentiations.

Exponential Function

Exponential functions are a special category of functions that involve constants raised to varying power expressions. These functions grow or decay rapidly. One fundamental property of exponential functions is that their derivative results in the function itself, sometimes scaled by a constant.

The standard exponential function can be represented as \(e^x\), where \(e\) is Euler's number, approximately 2.718. When considering a negative exponent, like \(e^{-x}\), the chain rule helps find the derivative.

In our problem, we consider \(e^{-x}\) as one of the functions. Upon differentiating, we get \(-e^{-x}\), illustrating how powerful exponential functions can be in both growth and decay models in various scientific fields.

Understanding the behavior of these functions and their derivatives crucially aids in solving problems in calculus, like the one presented here, where it leads to finding \(y(x) = C e^x\).