Let's start out with the definition of a derivative. We can use either form, but they both say the same thing:

lim(h-->0)[f(x+h)-f(x)]/h lim(x-->a)[f(x)-f(a)]/(x-a)

The second limit looks a lot like our slope formula: m = (y2 - y1)/(x2 - x1). So what exactly does this tell us? The derivative is the slope of a line tangent to a point. What happens is we calculate the slope between two points. As h-->0 (or x-->a), we grow infinitesimally closer to calculating [f(x)-f(x)]/(x-x).

Before we proceed to our shortcut rules for derivatives, I want to introduce notations first. We can either use the Newtonian notation or Leibnitzian notation. The Newtonian form is the f-prime form. For each derivative, we add an apostraphe to the fucntion. So the first derivative of f(x) is f'(x) (pronounced f-prime of x). The second derivative is f"(x), or f-double prime of x. Leibnitzian notation treats a derivative as a fraction, which will come in handy further down the road. For example, if we are differentiating y with respect to a variable x, we would have dy/dx as our first derivative. For an nth derivative, we would have d

^{n}y/dx

^{n}.

Below are a list of derivatives that you should commit to memory. You will need to know these like your multiplication tables for the rest of your Calculus career.

d/dx k = 0

d/dx x = 1

d/dx x

^{n}= n * x

^{n-1}

d/dx (f(x) + g(x)) = d/dx f(x) + d/dx g(x)

d/dx kf(x) = k * d/dx f(x)

d/dx sin(x) = cos(x)

d/dx cos(x) = -sin(x)

d/dx tan(x) = sec^2(x)

d/dx sec(x) = sec(x)tan(x)

d/dx csc(x) = -csc(x)cot(x)

d/dx cot(x) = -csc^2(x)

d/dx ln(x) = 1/x

d/dx log

_{b}(x) = 1/(x ln(B))

d/dx e^x = e^x

d/dx a^x = ln(a) * a^x

d/dx |u| = du/dx when to the right of the vertex, -du/dx when to the left of the vertex, and undefined at the vertex

**Power Rule**

This is one of the first rules listed above to memorize: d/dx x

^{n}= n * x

^{n-1}. So for example, if we have 3x^2, it's derivative is 6x. This is fairly straight-forward. We can also use this rule to quickly calculate a higher-order derivative, like say the 100th derivative, using factorials. As an example, let's say we have a function f(x) = 3x^150 + x^100 + 2x^20, and we ant to take the 100th derivative of f(x). To start, we know that the derivative of a constant is 0. Now b/c 20 < 100, we can simply treat 2x^20 as a constant. By the 20th derivative, only a constant will remain for that term. And on the 21st derivative, it will cancel out. For the term x^100, we can simply calculate 100! as the remaining term.

For our last term, 3x^150, we will have a power of 50 leftover after we take the 100th derivative. So we can calculate d

^{100}/dx

^{100}3x^150 by cancelling factorials. It comes out to 3 * 150!/(150-100)! * x^50.

Now simply add them together to get d

^{100}/dx

^{100}f(x) = 100! + 3(150!)/(50!) * x^50.

Or in a more general form, the rule is d

^{n}/dx

^{n}of x^r = r!/(r-n)! * x^(r-n).

**Product and Quotient Rules**

We use the product rule when calculating the derivative of a product of functions. With a constant, we can simply pull it out front and calculate the derivative of f(x). But when we try to d/dx f(x)g(x), we have to use a sepcial rule: d/dx f(x)g(x) = f(x)g'(x) + f'(x)g(x).

For example, if we have 3x^2 * ln(x), the derivative would be 3x^2/x + 6xln(x), which would simplify to 3x(1+2ln(x)). Notice how we go back to our derivative rule for ln(x).

When evaluating quotients, many people prefer to simply use the quotient rule: d/dx f(x)/g(x) = [g(x)f'(x) - f(x)g'(x)]/g^2(x). This is simply a special case of the product rule. Writing g(x) as g(x)^-1 will give us the same answer, and is one less rule to memorize.

For practice, let's find the derivative of sin(x)/x^3. By our quotient rule, that is [cos(x)x^3 - 3x^2sin(x)]/x^6. Rewriting 1/x^3 as x^-3, we still get the same answer: sin(x) * -3x^-4 + x^-3 *cos(x), just written a little bit differently.

**Chain Rule**

All of our derivative rules come together with the chain rule. Basically, the chain rule allows us to take derivatives of more complex functions like sin(cos^2(x)) or e^(3x^3). The rule is as follows: d/dx f(g(x)) = f'(g(x)) * g'(x). Or in short, we take the derivatives of the "inside functions" and multiple them by the derivative of the "outside function".

Let's look at our sin(cos^2(x)) function. To differentiate this, we need to take the derivative of the inside-most function: cos^2(x). Its derivative comes out to -2cos(x)sin(x). We use the power rule to get 2cos(x), then we multiple by the derivative of cos(x). Next, we take the derivative of sin(x), which is cos(x), and use the same parameter: cos^2(x). This leaves us with -2cos(x)sin(x)sin(cos^2(x)). Based on our trig identity, we can rewrite this as -sin(2x)sin(cos^2(x)).

Let's try our hand at the second example: d/dx e^(3x^3). We know that the derivative of e^x is e^x, so in the derivative function we will still have e^(3x^3). Now let's differentiate the 3x^3 term to get 9x^2. So our derivative is 9x^2 * e^(3x^3).

**Tangent Line Approximation**

Earlier in the tutorial, we said that the derivative of a function represents the slope of a tangent line at a point. So let's try our hand at calculating a tangent line for a function f(x) = 3x^3 + e^2x + ln(5x) at point x = 3.

We know our point-slope form looks like: (y-y1) = m(x-x1), with x1 = 3. So now let's find our y1 by plugging 3 into f(x) to get f(3) = 81 + e^6 + ln(30).

Next, let's find the derivative of f(x). We can determine that f'(x) = 9x^2 + 2e^2x + 5/x by our chain rule, as well as the other derivative rules listed above. From this, we get f'(3) = 81 + 2e^6 + 5/6. Since f'(3) is our slope, we simply plug the numbers into the formula:

y - (81 + e^6 + ln(30)) = (81 + 2e^6 + 5/6)(x-(81+e^6+ln(30))

**Conclusion**

This concludes my first tutorial. The next tutorial will cover applications of the derivative, interpreting what a derivative means in context, the first and second derivative tests, and L'Hospital's rule.