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Differentiation Shortcuts

By Edited Nov 13, 2013 0 0

As you probably know, finding a derivative is finding the slope of a tangent line. To do this, you must take a limit. The two forms of this limit are as follows:

Form One: m= limx→a(f(x)-f(a))÷(x-a)

This tells us the slope of the tangent line, 'm', is equal to the limit, as the variable 'x' approaches some number, 'a', of the function of 'x' minus the function of 'a', all divided by the difference between 'x' and 'a'.

Form Two: m= limh→0 (f(a+h)-f(a))÷h

This tells us that the slope of the tangent line, m'm, is equal to the limit, as the variable 'h' approaches the value zero, of the function of 'a+h' minus the function of 'a', all divided by 'h'.

To evaluate these, you must first know how to evaluate limits, and you must go through (what I believe to be) a long and arduous process, to come up with a final answer. When you first begin working these problems, it feels as if it takes forever to accomplish anything, and you probably are thinking " There must be a better way!" I know I was. On the bright side, if you are one of those people, you are in luck. There IS a better way!

This better way is a set of certain rules that work everytime when trying to find a derivative, if and only if, a function is differentiable. (Please note d/dx and 'D' are a differentiation operators, standing for taking the derivative; d/dx is Leibniz Notation.; and the Order of Operations Apply!) The rules are these:

1) Constant Function (Assuming 'C' is a constant): The derivative of ANY constant function is zero. d/dx (C)=0

Ex. d/dx (5)=0

2) The Power Rule (assuming 'n' is a positive integer): d/dx (xn)=nxn-1

Ex. d/dx (x3)=3x2

3) The Constant Multiple Rule (assuming 'c' is a constant and 'f' is a differentiable function): d/dx Cf(x)=Cd/dx f(x), the derivative of some constant times a function is equal to the constant times the derivative of the function.

Ex. d/dx (5·x3)= 5·d/dx (x3), both would be equal to 15x2

4) Sum Rule (If 'f' and 'g' are both differentiable): d/dx (f(x) + g(x))=d/dx f(x) + d/dx g(x), the derivative of two functions added together are equal to each of their respective derivatives added together.

Ex. d/dx (x2 + 2x2)= d/dx (x2) + d/dx (2x2), both would be equal to 6x.

5) Difference Rule(If 'f' and 'g' are both differentiable): d/dx (f(x)-g(x))=d/dx f(x)- d/dx g(x), the derivative on one function minus another, is euqal to the derivative of the first function minus the derivative of the second function.

Ex. d/dx (4x3-3x3)= d/dx (4x3)- d/dx (3x3), both would be equal to 3x2.

6) Product Rule (If 'f' and 'g' are both differentiable): d/dx f(x)·g(x)=f(x)·d/dx g(x) + g(x)·d/dx f(x), the derivative of two functions multiplied together is equal to the first function times the derivative of the second, plus the second function times the derivative of the first.

Ex. d/dx (x2)·(x3)= x2· d/dx (x3) + x3· d/dx (x2), both derivatives would be

equal to 5x4

7) Quotient Rule (If 'f' and 'g' are both differentiable): d/dx ((f(x)÷g(x))=(g(x)·d/dx f(x))- (f(x)· d/dx g(x))÷ (g(x))2,meaning the derivative of one

function divided by another is the first times the derivative of the second minus the quantity of the second minus the derivative of the first, all divided by the denominator of the original squared.

Ex. d/dx (x3)÷(x2)= ((x2)· d/dx (x3)-((x3)· d/dx (x2)))÷(x2)2, both

would be equal to 1.

The way I was tought to remember the quotient rule is to assign each the numerator and denominator their own name, "hi" for the numerater and "ho" for the denominator, so: d/dx of ho÷hi= hoDhi-hiDho/hoho, meaning the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator all divided by the denominator, squared.

As a side note, the quotient rule can be used to extend the earlier discussed power rule in the case where n is a negative integer. It would look like d/dx (x-n)= -n-n-1. This would work just as the power rule would in every aspect, just with negative numbers.

These are the derivative shortcuts that should cut down on your pain and agony in taking derivatives through evaluating limits.


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