Question

Using the Maclaurin series expansions for e x

,sin(x), and cos(x), show the following: 1. dx
d
(e x
)=e x
2. dx
d
(sin(x))=cos(x) 3. e ix
=cos(x)+isin(x), where i= −1
. Hint for part 3: notice that i 1
=i,i 2
=−1,i 3
=−i, and i 4
=1. In general, if r is the remainder when k is divided by 4 , then i k
=i r
, and r=0,1,2, or 3 .

Answer 1

Let's proceed step-by-step:

1. Find the derivative of e^x (with respect to x):
The derivative of a function measures how the function changes as the input (in this case 'x') changes. Specifically, for the function e^x, the rate of change is the same as the function value at that point. In other words, the slope of the tangent line to the function at any point is e^x. Hence the derivative of e^x with respect to x is itself, e^x.

2. Find the derivative of sin(x) (with respect to x):
The derivative of sin(x) is a standard result in calculus. We find that as we change 'x' slightly, the value of sin(x) changes approximately as much as cos(x) times the small change in 'x'. Hence the derivative of sin(x) with respect to x is cos(x).

3. Express e^(ix) as a combination of cos(x) and i*sin(x):
This is known as Euler's formula which states that for any real number 'x',
e^(ix) = cos(x) + i*sin(x)
This equation is proved using the Maclaurin series (or Taylor series centered at zero) for the exponential, sine, and cosine functions. Notice that e^(ix) gives a point on the unit circle in complex plane corresponding to angle x in radian.

In summary, we know:
The derivative of e^x is e^x.
The derivative of sin(x) is cos(x).
The expression e^(ix) expands to cos(x) + i*sin(x).

This is the result we anticipated.