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Jarrod Asuncion Period 1 Brose. Simple Derivatives Sample 1 ◊^n = something to the “something” power n ∙ ◊^n-1 ∙ d◊ = multiply something by ‘n’ and the.

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Presentation on theme: "Jarrod Asuncion Period 1 Brose. Simple Derivatives Sample 1 ◊^n = something to the “something” power n ∙ ◊^n-1 ∙ d◊ = multiply something by ‘n’ and the."— Presentation transcript:

1 Jarrod Asuncion Period 1 Brose

2 Simple Derivatives Sample 1 ◊^n = something to the “something” power n ∙ ◊^n-1 ∙ d◊ = multiply something by ‘n’ and the derivative of something; also, subtract 1 from the exponent. Ex. x^3 = 3x^(3-1) ∙ d(x) = 3x^2

3 Product Rule Sample 1 ◊∆ = 1something times 2something ◊’∆ + ◊∆’ = [derivative of 1something times 2something] plus [1something times derivative of 2something] Ex. x^3[((x^2)-1)^3] ◊=x^3∆= ((x^2)-1)^3 ◊’= 3x^2∆’= 3((x^2)-1)^2 ∙ 2x OR 6x((x^2)-1)^2 ◊’∆ + ◊∆’=[3x^2][((x^2)-1)^3] + [x^3][6x((x^2)-1)^2] = [3x^2][((x^2)-1)^3] + [6x^4((x^2)-1)^2] = 3x^2 ((x^2)-1)^2 ∙ [(3x^2)-1] -combine like terms

4 Quotient Rule Sample 1 ◊ ∕ ∆ = 1something divided by 2something (◊’∆ - ◊∆’) ∕ ∆^2 = [deriv. of 1 something times 2something] – [1something times deriv. of 2something], all over (2something)^2 Ex. (x^3) / ((x^2)-1)^3◊=x^3∆=((x^2)-1)^3 ◊’=3x^2 ∆’=6x((x^2)-1)^2 3x^2[((x^2)-1)^3] – x^3[6x((x^2)-1)^2] = (3x^2)((-x^2)-1) [((x^2)-1)^3]^2 ((x^2)-1)^3


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