Derivatives

Derivatives are the slope of a function at a particular point. You already know the slope of a line: (f(x2)-f(x))/(x2-x). If x2 = x+d, then (f(x2)-f(x))/(x2-x) = (f(x+d)-f(x))/(x+d-x) = (f(x+d)-f(x))/d. If you do that for a line, you get the same slope no matter which x and x2 you pick. If you do that for some other function f(), where f() is fairly smooth, then for any x, (f(x+d)-f(x))/d is always about the same value if d is small enough. That value is the slope of f() at x, and is also called the derivative of f() at x.

If you can calculate f(x), you can approximate the derivative with your calculator. Suppose the f(x) you are looking at is x³, and you want its derivative at 1. You can calculate that 1.1³ = 1.331, 1.01³ = 1.030301, 1.001³=1.003003001. And the corresponding slopes are f(1.1)-f(1)/.1 = 3.31, (f(1.01)-f(1))/.01 = 3.0301, (f(1.001)-f(1))/.001 = 3.003001. You can see the derivative at f(1) is probably 3.

One place derivatives are used in real life is in finding best answers: the minimum and maximum values of functions always have a slope of 0, meaning the derivatives there are 0. Solving for the roots of the derivatives tells you where to find the minimums and maximums of the original function. For example, the derivative of x³-3x is 3x²-3, so the minima and maxima of x³-3x are the roots of 3x²-3. Since 3x²-3 = 3(x+1)(x-1), the minima and maxima of x³-3x are at -1 and 1.

Derivatives and integrals of polynomials

Using algebra, you can find the exact derivative for any xⁿ. You know (x+d)ⁿ = xⁿ + nx^n-1d + (n choose 2)x^n-2d² + (n choose 3)x^n-3d³ et cetera. So the slope is going to be ((xⁿ + nx^n-1d + d²(stuff)) - xⁿ)/d. The two xⁿ cancel out. A d cancels out. That leaves nx^n-1 + d*(stuff). For example the derivative of x³ is 3x² + d*(3x + d). If d is small enough (a millionth, a trillionth), "stuff" can't matter, so you're left with nx^n-1. The derivative of xⁿ is nx^n-1. That's the most important thing in calculus, at least half of its total worth to the rest of your life, so remember that.

An integral is the opposite of a derivative: since the derivative of x³ is 3x², that tells you that the integral of 3x² is x³. (Plus a constant. Since the derivative of every constant is 0, the integral of 0 can be any constant.) What's the integral of x²? Using algebra, it's x³/3 (plus a constant). The integral of xⁿ is xⁿ⁺¹/(n+1) (plus a constant). The derivative of the integral of f(x) is f(x). The integral of the derivative of f(x) is f(x) (plus a constant).

A polynomial is a sum of powers of x, like x³ + 4x² + 3. The derivative of the whole is the sum of the derivatives of each of the terms, like in this case 3x² + 8x + 0. The integral of the whole is the sum of the integrals of each of the terms (plus a constant).

The reason this is so useful is that any smooth function can be split into pieces, and you can approximate each piece with a polynomial. The derivative of the function is then approximated by the derivatives of the polynomials for each piece, and the integral of the function is approximated by the integrals of the polynomials for each piece. Since you know how to do derivatives and integrals of polynomials, you know how to approximate derivates and integrals of everything.

e

Define e(x) as the function where e(0) = 1 and the derivative at e(x) is e(x).You recall e^x? It turns out e(x) is e^x, the derivative of e^x is e^x, and the integral of e^x is e^x (plus a constant).

You can approximate any function with polynomials, so let's approximate e(x) near 0 with some polynomial. What's that polynomial?

• e(0) = 1, by definition.
• Since the derivative of e(x) = e(x), the derivative of the first term has to be 1, so the first term has to be x.
• Since the derivative of e(x) = e(x), the derivative of the second term has to be x, so the second term has to be x²/2.
• Since the derivative of e(x) = e(x), the derivative of the third term has to be x²/2, so the third term has to be x³/6.
• Recursively, since the n-1th term of e(x) is x^n-1/(n-1)!, the nth term has to be xⁿ/(n!).

That tells you that e(x) = 1 + x + x²/2 + x³/6 + x⁴/24 + x⁵/120 + x⁶/760 + x⁷/5040 + etc. If e(x) really is an exponential, it also defines e as e = e¹ = e(1) = 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120 + 1/760 + 1/5040 + etc. That happens to come out to e = 2.71828... . (Todo: show e(x) is actually e^x.) This polynomial approximation is how computers really go about calculating e^x: they have to keep adding terms to the sum until the remaining terms are smaller than the desired precision. The later terms always get small enough eventually. That's because n! eventually grows faster than xⁿ. But terms get small fastest when x is near 0. So computers use more math to make it go faster, for example they use e^x+1 = e*e^x to choose an x near 0.

Complex numbers, Cosine, and Sine

You know the real numbers, like -0.123, and square roots. And you know there is no real-number square root of negative numbers.

Define i to be the square root of -1. There is no real-number square root of -1, so i is some other type of number, call it an imaginary number. i² = -1, i³ = -i, and i⁴ = 1. Multiplying i by a real-numbered coefficient gives you more imaginary numbers, like -0.123i. A complex number is any sum of a real number and imaginary number, for example 1.2, -3.4i, or 5 + 6.7i. You can draw complex numbers on a graph, using the real component for x (horizontal) and the imaginary component for y (vertical).

What is the polynomial approximation of e^bi? Remember that i² = -1, and plug bi into the polynomial appromation of e(x): 1 + bi - b²/2 - b³i/6 + b⁴/24 + b⁵i/120 - b⁶/720 - b⁷i/5040 + etc.

You can define sin(b) and cos(b) (b in radians) this way: e^bi = cos(b) + isin(b). (So far sine and cosine are just some defined functions.) If you plot e^bi on a graph, it forms a circle of radius 1 around the origin, with cos(b) as the real coordinate and sin(b) as the imaginary coordinate. π is defined as the smallest positive real number where e^πi = -1. That happens to come out to π = 3.14159... . Note e^πi/2 = i, e^πi = -1, e^3πi/2 = -i, and e^2πi = 1, matching what the powers of i should be.

You can separate the terms of e^bi into imaginary numbers (multiplied by i) and real numbers (not multiplied by i): e^bi = (1 - b²/2 + b⁴/24 - b⁶/720) + ...) + i(b - b³/6 + b⁵/120 - b⁷/5040 + ...). Every even term is real, every odd term is imaginary. This gives you a way to calculate sine and cosine with polynomials, rather than using the trigonometry buttons on your calculator:

• cos(b) = 1 - b²/2 + b⁴/24 - b⁶/720 + ...
• sin(b) = b - b³/6 + b⁵120 - b⁷/5040 + ...

It also tells you what the derivative of sine and cosine are. Look at the polynomial approximation of sine, take its derivative, and compare the result to the polynomial approximation of cosine. The derivative of sin(x) is cos(x), and the derivative of cos(x) is -sin(x). This isn't a coincidence: it has to be that way because e(x) was defined based on being its own derivative.

Exponentials of complex numbers are always complex numbers: e^a+bi = e^ae^bi = e^a(cos(b)+isin(b)).

You can square those series above to see what cos²(b) and sin²(b) actually are:

• cos²(b) = 1 - b² + b⁴/3 - 2b⁶/45 + ...
• sin²(b) = 0 + b² - b⁴/3 + 2b⁶/45 - ...