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by: Megan steltz

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# Product, Quotient Rule, Trigonometric Derivatives Math 120-4

Marketplace > Eastern Michigan University > Math > Math 120-4 > Product Quotient Rule Trigonometric Derivatives
Megan steltz
EMU
GPA 3.28

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product rule quotient rule trigonometric derivatives also contain proofs for the product and quotient rule
COURSE
Calculus I
PROF.
John G. Patrick
TYPE
Class Notes
PAGES
6
WORDS
CONCEPTS
Math, Calculus, Product, Derivatives calculus
KARMA
25 ?

## Popular in Math

This 6 page Class Notes was uploaded by Megan steltz on Sunday October 9, 2016. The Class Notes belongs to Math 120-4 at Eastern Michigan University taught by John G. Patrick in Fall 2016. Since its upload, it has received 4 views. For similar materials see Calculus I in Math at Eastern Michigan University.

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Date Created: 10/09/16
Product Rule If F(x)=f(x)g(x), Then F’(x)=f(x)g’(x) + f’(x)g(x) Ex. d/dx [ (4x^3 - x^2 -1)(x^3 -2x^2 + 3x +1)] = (4x^3 - X^2 - 1) (3x^2- 4x + 3) = (12x^2 -2x)(x^3 - 2x^2 + 3x + 1) However the product rule is more useful for the product of different types of functions Ex. f(x)= x^2cos(x) g(x) = e^x arctan(x) Proof of the Product Rule Let F(x) = f(x)g(x) F’(x) = lim ((F(x + h ) - F(x))/h) h 0 =lim ((f(x + h )g( x + h) - f(x)g(x))/h) h 0 Subtract and add f(x + h)g(x) = lim ((f( x + h)g(x + h) - f(x + h)g(x) + f(x + h) - g(x) - f(x)g(x))/h) h 0 = lim ((f( x + h)(g( x + h) - g(x)) + g(x)(f( x + h) - f(x))/h) h 0 = lim ((f ( x+ h)(g( x + h) - g(x))/h) + (g(x)f(x + h) - f(x)/h) h 0 = lim (f( x + h) lim ((g(x + h) - g(x))/h) + lim (g(x))lim((f(x + h) - f(x))/h) h 0 h 0 h 0 h 0 = f(x)g’(x) + g(x)f’(x) The Quotient Rule d/dx (f(x)/g(x)) = g(x)d/dx f(x) - f(x)d/dx (g(x))/(g(x))^2 If F(x) = f(x)/g(​ ’(x) = (g(x)f’(x) - f(x)g’(x))/(g(x)^2) Rhyme for remembering quotient rule “ Low times d-high minus high. D-low, over the square of what’s below” Ex. d/dx [x/x^2-1] = ((x^2-1)(1) - x(2x))/(x^2-1)^2 = (x^2 - 1 - 2x^2)/ ( x^4 - 2x^2 + 1) = (- x^2 - 1)/ ( X^4 - 2x^2 + 1) d/dx( x^2 + 2x)/(x^4 - 3x^2 + 1) = ((x^4 - 3x^2 +1)(2x + 2) - (x^2 + 2x)(4x^3 - 6x))/(x^4 - 3x^2 + 1)^2 Graph Note. ​with a function like f(x) = 3/ = 3/x^½ It is easier to rewrite it a 3x^-½ and use the power rule =f’(x) = -(3/2)x^-3/2 Using the quotient rule F’(x) = (x^(½)(0) - 3(½)x^-½)/(x^½) d/dx(x^½(1-x^5) The answer for the power rule and the quotient rule are the same Derivatives of Trigonometric Functions At x = -3π/2, − π/2, π/2, 3π/2, ...., f’(x) must be zero On the interval ( − π/2, π/2,) sin(x) is increasing so f’(x) is positive On (π/2, 3π/2) f’(x) must be negative It looks like d/dx sin(x) = cos(x) By the same method we could guess that d/dx(cos(x))= -sin(x) Thm ​d/dx(sin(x)) = cos(x) Proof d/dx(sin(x)) = lim ((sin( x + h) - sin(x))/h) h 0 Using the addition formula for sin(x) = lim (sin(x)cos(h) + cos(x)sin(h) - sin(x))/h) h 0 = lim ((sin(x)cos(h)-sin(x))/h) + (cos(x)sin(h))/h) h 0 =lim((sin(x)cos(h)-1)/h) + cos(sin(h)/(h) h 0 = lim (sin(x)lim((cos(h)-1)/h) } this is equal to zero h 0 h 0 lim(cos(x))lim(sin(h)/h) } this is equal to one h 0 h 0 = sin(x)*0 = cos(x) * 1 =cos(x) A mass on the end of a spring is pulled down 1 foot and released at time = 0. It moves in s ​ imple harmonic motion Its displacement at time t is given by s(t)= -cos(t) It velocity at time t is given by v(t) = s’(t) = sint(t) t 0 π/2 π 3π/2 2π s(t) -1 0 1 0 -1 v(t) 0 1 0 -1 0 To find d/dx (tan(x)), we use the quotient rule d/dx (tan(x)) = d/dx(sin(x)/cos(x)) =(cos(x)d/dx sin(x)-sin(x)d/dx cos(x))/(cos(x))^2 = (cos(x)cos(x) - sin(x)-sin(x))/cos^2(x) = (cos^2(x) + sin^2(x))/cos^2(x) = 1/cos^2(x) = 1/cos^2(x) = sec^2(x) Graph of tan(x) graph of sec^2(x) To find d/dx sec(x), use the quotient rule d/dx sec(x) = d/dx(1/cos(x)) = (cos(x)(d/dx(1)) - 1*d/dx(cos(x)))(cos(x)^2) = (cos(x)*0 -1 *(-sin(x))/cos^2(x) = (sin(x))/(cos^2(x) = (1/(cos(x))* (sin(x))/(cos(x) = sec(x)tan(x) Ex. d/dx(tan(x)/x^2 = (x^2 d/dx(tan(x))-tan(x)d/dx(x^2))/(x^2)^2 = (x^2sec^2(x)-tan(x)*2x)x^4 =(xsec^2(x)-2tan(x))x^3

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