Derivative+Techniques

=Derivative Techniques - Jake and Michael=


 * Power rule

The derivative of x to a power is the power times x to a power one less than the original power.

example 1:

f(x) = x -2, then f '(x) = -2 x -3 = -2 / x 3

as a short proof:

power rule is true when //n// = 1: //dx// || //x// 1 || =1**·** //x// 0  = 1; || //dx// || //x// 2 || = 2//x//; || //dx// || //x// 3 || = 3//x//². ||
 * ||  || __//d//__
 * that it is true when //n// = 2: ||
 * ||  || __//d//__
 * ||  || __//d//__
 * ||  || __//d//__
 * and that it is true when //n// = 3: ||
 * ||  || __//d//__
 * ||  || __//d//__
 * ||  || __//d//__

proof by induction: The induction hypothesis will be that the power rule is true for //n// = //k//: //dx// || //x// //k// || = //k// //x// //k//−1 , || and we must show that it is true for //n// = //k// + 1; i.e. that //dx// || //x// //k//+1 || = (//k// + 1) //x// //k//. || Now, //dx// || //x// //k//+1 || = || __//d//__ //dx// || //x//**·** //x// //k// ||
 * __//d//__
 * __//d//__
 * __//d//__

//dx// || //x// //k//+1 || = |||| //x//**·** //k x// //k−1//  + //x// //k//  **·** 1, according to the product rule,x || //dx// || //x// //k//+1 || = |||| //k x// //k//  + //x// <span style="font-family: garamond,serif;">//k//  || //dx// || //x// <span style="font-family: garamond,serif;">//k//+1 || = |||| (//k// + 1)//x// <span style="font-family: garamond,serif;">//k//. || <span style="color: #000066; font-family: garamond,serif; font-size: 20px; line-height: normal;">Therefore, if the power rule is true for //n// = //k//, then it is also true for its succesor, //k// + 1. And since the rule is true for //n// = 1, it is therefore true for every natural number.
 * __//d//__
 * __//d//__
 * __//d//__
 * __//d//__
 * __//d//__

<span style="font-family: Verdana,Geneva,sans-serif; font-size: 90%;"> media type="youtube" key="6kScLENCXLg" height="385" width="480" The derivative of a composite function is the derivative of the inside function times the derivative of the outside function (keeping the inside part intact).
 * <span style="font-family: Verdana,Geneva,sans-serif; font-size: 90%;">Chain Rule

<span style="font-family: Verdana,Geneva,sans-serif;"> The derivative of two multiplied functions is the addition of the products of one function and the other functions derivative, and vice versa.
 * <span style="font-family: Verdana,Geneva,sans-serif; font-size: 90%;">Product Rule

Proof:

 To prove the product rule, we will express the difference quotient Δ//x// ||. Thus let || //y// = //f g//. Then a change in //y// -- Δ//y// -- will produce corresponding changes in //f// and//g//: //y// + Δ//y// = (//f// + Δ//f// )(//g// + Δ//g//) On multiplying out the right-hand side, //y// + Δ//y// = //f g// + //f// Δ//g// + //g// Δ//f// + Δ//f// Δ//g//. Δ//x// ||, we subtract //y// -- which is //f g// -- and divide by Δ//x//: || Δ//x// || = //f// || __Δ//g//__ Δ//x// || + //g// || __Δ//f//__ Δ//x// || + Δ//f// || __Δ//g//__ Δ//x// || Now let Δ//x//0. Hence Δ//f// will approach 0 --  Therefore, //dx//** || **= //f//** || **__//dg//__ //dx//** || **+ //g//** || **__//df//__ //dx//** ||
 * simply as || __Δ//y//__
 * To form || __Δ//y//__
 * __Δ//y//__
 * **__//dy//__

example: f(x) = sinxcosx

f'(x) = (f')(g) + (g')(f) when, f = sinx f' = cosxg = cosx g' = -sinx f'(x) = (cosx)(cosx) + (sinx)(-sinx) = cos2x - sin2x dx

<span style="font-family: Verdana,Geneva,sans-serif; font-size: 90%;">
 * <span style="font-family: Verdana,Geneva,sans-serif; font-size: 90%;">Quotient Rule

The derivative of two functions divided is the subtraction of the product of the top function and the bottom function's derivative from the product of the top function's derivative and the bottom function. Then one divides that result by the square of the bottom function.

example: f(x) = (6x)/(x2 - 4) f'(x) = h(x) = [(f')(g) - (g')(f)]/(g)2 where, f = 6x f' = 6g = x2 - 4 g' = 2x h(x) = [(6)(x2 - 4) - (6x)(2x)]/(x2 - 4)2 = (6x2 - 24) - (12x2)/(x2 - 4)2

= (-6x2 - 24)/(x2 - 4)2

<span style="font-family: Verdana,Geneva,sans-serif;"> (d/dx) = f(x) The derivative of a function in an integral is just the function.
 * <span style="font-family: Verdana,Geneva,sans-serif; font-size: 90%;">Deriving an integral function

<span style="font-family: Verdana,Geneva,sans-serif; font-size: 90%;">First we consider that the derivative of lnx = 1/x. Second we consider logarithm rules. Any logarithm with any base can be rewritten via change of base formula... where b is any constant.
 * <span style="font-family: Verdana,Geneva,sans-serif; font-size: 90%;">Logarithmic differentiation

Third we consider that lnx is merely a logarithm with base e. It follows that we can rewrite any logarithm to contain only ln, which we know how to derive. We will always have an variable ln on top divided by a constant on the bottom. (d/dx)log_a(x) = (d/dx)lnx/ln(a) = (1/x)/lna = 1/(xlna)

Ex: (d/dx)log_5(x) = (d/dx)lnx/ln(5) = 1/(xln5)

sources:

[] [|http://www.math.hmc.edu/calculus/tutorials/quotient_rule/www.youtube.com]

Topic: Derivative Techniques Group Members: Jake and Michael

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Very thorough material with proofs included. Some of the notation isn’t great. Shouldn’t use ^2 for example. Mostly quotient rule section. No images or graphs to support your algebraic work. Creativity is lacking, but well organized page.

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