KNIGHTS OF THE EASTERN CALCULUS.
First Challenger
	Power Rule of Differentiation
	Sum and Difference Rules of Differentiation
	Constant Rule of Differentiation
Second Challenger
	Chain Rule of Differentiation
Third Challenger
	Product Rule of Differentiation
	Finding Product of Two Binomials (FOIL method)
Fourth Challenger
	Quotient Rule of Differentiation
	Finding Square of a Trinomial
Exercises Set 1

"I'm better than all you bozos!"

A cold wind blew through the street in front of Lain's house as she heard the electric wires humming their perpetual hum. The Knights had finally come!

After what seemed like an eternity of waiting the first challenger stepped forward. His question:
Power Rule of Differentiation.
"An object moves along a line. Its displacement s from O, a point on the line, is given by s = 5t^2 + 10t + 7, where t is time, in hours. What is its velocity v, in terms of t?"

Lain pondered for a moment. Her answer:
Sum and Difference Rules of Differentiation.
"To answer this question, we have to find the derivative of the function s(t) which is written s'(t). A derivative is an instantaneous rate of change and the process of finding it is called differentiation. The important thing about derivatives, is you always differentiate term by term; (a+b)' = (a)' + (b)'. First we look at 5t^2. To differentiate this term, we divide by t (or we reduce the exponent of the variable by one) and we multiply by the power of t (that is, the power it had before we reduced it by one). So first, we divide by t. Now we have t^1, which is just plain t. Then, we multiply 5 by 2, the power of t. We now have 10t. That's our first term.

"For the second term, dividing t by t gives t^0, which, for every number, is 1. And multiplying 10 by 1 is just 10. So we have 10. That's our second term.
Constant Rule of Differentiation.
"For the third term, there is no variable. Hence, it will stay the same whatever we make t. So, there is no change, and therefore no rate of change. So we have 0. That's our third term.

"So the answer is v = 10t + 10.

"Idiot," she stared into his eyes and he dissolved into nothingness from the force of her glare.

The second challenger stepped forward, supremely confident. His question:
Chain Rule of Differentiation.
"Differentiate (x + 1)^100. Let's see how long you'll take to open up this bracket! Ha ha ha ha ha!"

Lain remained unperturbed. Her answer:
"I don't have to. Notice that (x + 1)^100 has two steps. The first is (x + 1) and the second is the result of the first step to the power of 100. What we do now is differentiate each of those steps and multiply the derivatives. First, we differentiate (x + 1). The answer is 1.

"Next, we substitute u for (x + 1), so u = x + 1. Then we differentiate u^100; the answer is 100u^99. Now, we multiply the derivatives and undo the substitution of u. So we have 1(100u^99) = 100(x + 1)^99.

"Idiot," she stared into his eyes and he dissolved into nothingness from the force of her glare.

A third challenger appeared. He was angry and determined to avenge his companions. His question:
Product Rule of Differentiation.
"What is the velocity v of a car whose displacement s with respect to time t is modeled by the function s = (t^2 + 5t + 2)(t + 3)? Try factoring out this!"

Lain was cool and collected. Her answer:
"The derivative of the product of two functions is the sum of the products of the derivative of each function and the original of the other. In math terms, let h(x) be the product of two functions f(x) and g(x). Then h'(x) = f'(x)g(x) + f(x)g'(x).

"Usually we call the two parts u and v. Substitute u for t^2 + 5t + 2 and v for t + 3. Then the answer is u'(t)v(t) + v'(t)u(t).

"We know that u' equals 2t + 5 and that v' equals 1. So the answer is (2t + 5)(t + 3) + (1)(t^2 + 5t + 2). That reduces to (2t + 5)(t + 3). Now we use FOIL to open this one up. The product of the First terms, 2t and t, is 2t^2. The product of the Outer terms, 2t and 3, is 6t. The product of the Inner terms, 5 and t, is 5t. The product of the Last terms, 5 and 3, is 15. 6t plus 5t is 11t. So the answer is v = 2t^2 + 11t + 15. This v is not the same v as we used in solving this problem."

"Idiot," she stared into his eyes and he dissolved into nothingness from the force of her glare.

A fourth challenger appeared. His question:
Quotient Rule of Differentiation.
"Differentiate (3x + 7)/(2x^2 + 5x + 6)."

Lain thought and had the answer. Her answer:
"The derivative of a fraction h(x) with numerator f(x) and denominator g(x) is the derivative of the numerator times the denominator, minus the derivative of the denominator times the numerator, all divided by the square of the denominator. So h'(x) = (f'(x)g(x) - f(x)g'(x))/(g(x))^2.

"Like what we did above, we call f(x) u and g(x) v. So u is (3x + 7); this makes u' simply 3. And v is (2x^2 + 5x + 6); by what we have learned, v' is (4x + 5). Now we know that the answer is ((3)(2x^2 + 5x + 6) - (4x + 5)(3x + 7))/(2x^2 + 5x + 6)^2.

"Expanding, we have ((6x^2 + 15x + 18) - (12x^2 + 15x + 28x + 35))/(2x^2 + 5x + 6)^2. Remembering to change sign, and combining like terms, we have (-6x^2 - 28x - 17)/(2x^2 + 5x + 6)^2.

"If you're still not satisfied we can expand the denominator. (4x^4 + 10x^3 + 12x^2 + 10x^3 + 25x^2 + 30x + 12x^2 + 30x + 36) = (4x^4 + 20x^3 + 49x^2 + 60x + 36).

"So our final answer is: (-6x^2 - 28x - 17)/(4x^4 + 20x^3 + 49x^2 + 60x + 36).

"You may have wasted my time but you're still an idiot," she stared into his eyes and he dissolved into nothingness from the force of her glare.

"Hey, you!" Lain's finger pointed out of the screen. "Yes, you! I can't hold off the Knights all by myself! Do these exercises and do your part!"
1. Differentiate x^4 + 3x^3 + 7x^2 + 12x + 26.
2. Differentiate 3/x. (Hint: Represent 1/x as x^-1).
3. Differentiate (x + 7)(x - 5).
4. Differentiate (x^2 + 5x + 6)/(x + 3).
5. Find the rate of change at time t of a quantity q modeled by q = (3x^3 + 2x^2 + x + 5)/(x + 2).
6. Differentiate cos(2x). (Note: The derivative of cos(x) is -sin(x).)
7. The population p of a colony of bacteria at time t is represented by p = (6t)^7. What function can we use to find the rate of change of p at time t?