Derivaties And Integrals

When discussing derivatives and integrals, it is always important to know where

you are in relation to the function. If you were on a staircase, and you had the original

function [f(x)], you would be in the middle step. If you were then to take the derivative of

that function [f Ð''(x)] you would need to take a step down. Going back up to the original

function, if from there you were take a step up, you would have found the integral of that

function [ f(x)] {billy, draw in an integral sign back there!!!}

Integral of f(x)

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Original Funct. f(x)

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[

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Derivative of f(x)

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[

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{billy under the step of der write " f Ð''(x) = (x)^n-1" and then under orig funct write "f(x) = (x)^n" and then under integ write " S (x) = (x)^(n+1) + C"}

If you are given a function and asked to find the derivative, you need to "take a step

down". You'll take all your x^n's to one less degree. But that's not all. With derivatives,

you must also remember to multiply the new x's BY n. For example, with a function like

x^3, you not only have to reduce the n by one (now x^2) but you need to swing the

original n (3) out and multiply by it. x^3 now becomes 3x^2. Even if there is a coefficient

k, you still need to multiply by n. Hence, the derivative of a function like 4x^5 you be

5 X 4x^4, or 20x^4. {you should go back and delete the x^2 ex. And write the actual "x

squared, cubed", etc} When there's simply and x, not raised to anything, the power n is

assumed to be one. So the derivative of x^1 = x^(1-1) = x^0 = 1. The rule with

coefficients still applies, so the derivative of the function x^3 + 4x would be 3x^2 +4(1) =

3x^2 +4.

Crossing constants is very common when solving derivatives, and you need to know how

to deal with them as well. With derivatives, the solution is very simple. The derivative of

a constant c is always zero. f Ð''(19) = 0 and f'(2x^2 + 5x Ð'-