How to get the derivative of a function

By definition: Initially, in most calculus I courses the student is required to compute the derivative by use of it's definition. In words, it would be the limit of the secant line slopes as the change in x between the two points approach 0. This does not help us get it. We need a workable mathematical way of getting it. So, the following is what this looks like:

dy/dx = lim Δy/Δx
        Δx→0

Note: In many cases it's more convenient to use the functional notation equivalent of the above to calculate the derivative. In that case, we use a new symbol, f '(x) for dy/dx and one letter h for Δx and f(x+h)-f(x) for Δy. It will have the following look:

  f '(x) = lim [f(x+h)-f(x)]/h

h→0


There is a third form for the derivative that was used in the previous section that concentrates on getting the derivative at a fixed point [a,f(a)] by letting the variable point [x,f(x)] approach the fixed point on the curve. It has the following form:

 i.e.,      f '(a) = lim[f(x)-f(a)]/(x-a)                                                       
                                                                                                                                     x→a


Example of using the limit definition of the derivative to find its formula

Let's take the function f(x) = -2x2-3x-5. We wiil formulate the derivative in steps using the above definition.

1) f(x+h)= -2(x+h)2-3(x+h)-5 (just replace x by x+h where you see an x)

= -2(x2+2hx+h2) -3(x+h)-5 (square the x+h)


= -2x2-4hx-2h2-3x-3h-5 (eliminate the parentheses)

2) f(x+h)-f(x)= -2x2-4hx-2h2-3x-3h-5-(-2x2-3x-5) (subtract the f(x) expression)
= -2x2-4hx-2h2-3x-3h-5+2x2+3x+5 (eliminate the parentheses)

= -4hx-2h2-3h (simplify) all terms should contain a least one factor of h)

3) [f(x+h)-f(x)]/h =-4x-2h-3 (divide each term by h)

4) lim [f(x+h)-f(x)]/h h->0 = -4x-3 (sub in 0 for h) all terms with h will disappear)

this is f '(x)=-4x-3 (this is your answer)

Not all derivatives are this easy. Some require quite a bit of algebra before the division in step 3

Note: Since we need to calculate many derivatives in calculus, this process can be time-consuming every time we need a derivative. So, short-cuts are developed that will enable us to quickly compute them. Usually, calculating the derivative by definition is only used at the very beginning during the development of the derivative for theorical purposes. Short-cuts are coming up.


DERIVATIVE FORMULAS

Note: In a typical calculus I course, these formulas are proven using the basic definition of the derivative. Once proven, getting the derivative by definition is rarely used and often forgotten. I've included a comprehensive list of formulas that involve different classes of functions. Many of you many not be familar with these and will need to study much of pre-calculus in order to feel comfortable working with them. During our application discussion (how to use the derivative to solve basic problems), we will confine our discussion to the very basic functions. However, realize that persons in the scientific world deal with most all of these functions on a continuous basis.

Constant: If y = k, any constant, then dy/dx = 0

Example: If f(x)=7, then f '(x)=0

Power Rule: If y=un, then dy/dx = n u(n-1)du/dx

Example: If g(t)= (t2 + t - 5)7, then g '(t)= 7(t2+t-5)6 (2t+1)

General Expo: If y = au, then dy/dx = au (ln a) du/dx

Example: If y=2(x2), then y'=[2(x2)](ln2)(2x)

The special Expo: If y = eu, then dy/dx = eu du/dx

Example: If C=e(1-2t), then C ' = [e(1-2t)](-2)

The natural log: If y = ln(u), then dy/dx = (1/u) du/dx

Example: If P(x)= ln(x2-4x+33), then P '(x)= [1/(x2-4x+33)] (2x-4)

The trig fcns: If y = sin(u), then dy/dx = cos(u) du/dx

If y = cos(u), then dy/dx = - sin(u) du/dx

If y = tan(u), then dy/dx = [sec2(u)] du/dx

If y = sec(u), then dy/dx = sec(u)tan(u) du/dx

If y = csc(u), then dy/dx = - csc(u)cot(u) du/dx

If y = cot(u), then dy/dx = [- csc2(u)] du/dx

Sums: If y = u + v, then dy/dx = du/dx + dv/dx

Product Rule: If y = uv, then dy/dx = u dv/dx + v du/dx

Quotient Rule: If y = u/v, then dy/dx = [v du/dx - u dv/dx] / v2

Note: I've eliminated many other types of functions along with advanced techniques used for the more complicated types. A formal calculus course would cover them in dept. The best way to remember these is to do many problems dealing with them. The more you do, the easier it will be to recall them.