Understanding the Derivative of eix: A Comprehensive Guide

When it comes to understanding the derivative of complex functions in mathematics, one common question is the derivative of eix. This article aims to explore the derivative of this function using the chain rule, explain the process in simple terms, and provide a general understanding of the derivative of efx for any differentiable function f(x). Additionally, the article will also delve into the power series representation of the exponential function and its unique property of being its own derivative.

Derivative of eix

Let's begin by defining the derivative of eix, where is the imaginary unit. The derivative of eu with respect to x is eu middot; du/dx. Let's set u ix:

First, differentiate u: du/dx i. Apply the chain rule: d/dx eix eix middot; i. Hence, the derivative of eix is: d/dx eix ieix.

Simple Case and Generalization

For a simple case, treat i as a constant, then dy/dx i eix. We can generalize this concept for any differentiable function f(x). Let's now explore the derivative of efx.

Derivative of eax

The derivative of eax is aeax. Therefore, d/dx [eix] i.eix.

General Derivative of efx

To find d/dx efx, observe that fx ln(efx). Taking the derivative with respect to x of both sides:

Let efx g(x). Then fx ln(g(x)). Using the chain rule: d/dx ln(g(x)) 1/g(x) middot; g'(x). Since g(x) efx, d/dx efx fx middot; efx.

Power Series Expansion of ex

The exponential function ex can be represented as a power series:

ex 1 x x2/2! x3/3! x4/4! sum from k0 to infinity {xk/k!}

To differentiate this series term by term:

dy/dx ex 0 1 2x/2! 3x2/3! 4x3/4! 1 x x2/2! x3/3! ex

This shows that ex is equal to its own derivative at all points! In fact, it's the only function (up to a constant multiple) that has this property.

Proof of the Uniqueness Property

Suppose a function f(x) is equal to its own derivative, i.e., f(x) f'(x). Let g(x) e-x f(x). Using the chain rule for differentiation:

f'(x) -e-x f(x) e-x f'(x) e-x [f(x) - f'(x)]

Since f(x) f'(x), the expression simplifies to:

f'(x) e-x [f(x) - f(x)] 0

This means g(x) is just a constant, say g(x) k. Therefore, f(x) kex for some constant k.