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Download Useful Formulae Exam 1 - Iowa State University
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Leader: Wheaton Useful Formulae Course: Math 165 Supplemental Instruction Instructor: Dr. Tokorcheck Iowa State University Date: 02/11/15 Below is a list of formulae that it will be helpful to know moving forward in calculus that we have learned or needed to know to this point. While this list is by no means exhaustive, it should be a useful starting point. Trigonometric Formulae Limit Laws lim π = π Angle Sum Identities π₯βπ sin(πΌ + π½) = sin(πΌ) cos(π½) + cos(πΌ) sinβ‘(π½) lim π₯ = π π₯βπ cos(πΌ + π½) = cos(πΌ) cos(π½) β sin(πΌ) sinβ‘(π½) πππ 2 (π₯) + π ππ2 (π₯) = 1 1 + π‘ππ π₯βπ lim[π(π₯) + π(π₯)] = lim π(π₯) + lim π(π₯) Pythagorian Identities 2 (π₯) lim ππ(π₯) = π β lim π(π₯) π₯βπ 2 = sec (π₯) 1 + πππ‘ 2 (π₯) = csc 2 (π₯) Odd-Even Identities sin(βπ₯) = β sin(π₯) cos(βπ₯) = cos(π₯) tan(βπ₯) = β tan(π₯) π₯βπ π₯βπ π₯βπ lim[π(π₯) β π(π₯)] = lim π(π₯) β lim π(π₯) π₯βπ π₯βπ π₯βπ lim[π(π₯) β π(π₯)] = lim π(π₯) β lim π(π₯) π₯βπ π₯βπ lim [ π₯βπ π₯βπ lim π(π₯) π(π₯) ] = π₯βπ π(π₯) lim π(π₯) π₯βπ lim[π(π₯)]π = [lim π(π₯)] π₯βπ π π₯βπ π lim βπ(π₯) = πβlim π(π₯) π₯βπ π₯βπ Continuity Differentiable For a function to be continuous it must satisfy three conditions: A function π(π₯) is differentiable if πβ²(π) exists for every π in a given interval. 1) π is in the domain of π 2) lim π(π₯) β‘ππ₯ππ π‘π π₯βπ 3) lim π(π₯) = π(π) An infinitely differentiable function is called smooth (examples include π ππ(π₯) and πππ (π₯)) π₯βπ Derivative Definitions Derivative Rules π(π + β) β π(π) ββ0 β This definition is useful to evaluate the derivative at a given point π. Power Rule πβ²(π) = lim π π (π₯ ) = ππ₯ πβ1 ππ₯ π(π₯ + β) β π(π₯) ββ0 β π β² (π₯) = lim 1060 Hixson-Lied Student Success Center οΆ 515-294-6624 οΆ [email protected] οΆ http://www.si.iastate.edu This definition gives a function that represents derivative of the function π(π₯) π(π₯) β π(π) π₯βπ π₯βπ This definition is useful to calculate inverse functions. π β² (π₯) = lim Trig Derivatives π (cos(π₯)) = β sin(π₯) ππ₯ π (sin(π₯)) = cos(π₯) ππ₯ π (tan(π₯)) = sec 2 (π₯) ππ₯ π (cot(π₯)) = β csc 2 (π₯) ππ₯ π (secβ‘(x)) = tan(π₯) sec(π₯) ππ₯ π (cscβ‘(x)) = cot(π₯) csc(π₯) ππ₯ Squeeze Theorem Limits lim ( π₯β0 1 β πππ (π₯) )=0 π₯ π ππ(π₯) lim ( )=1 π₯β0 π₯ Common Exponential Derivatives π π₯ (π ) = π π₯ ππ₯ π π₯ (π ) = ln(π) β π π₯ ππ₯ Linearity of Derivatives π π (π β π(π₯)) = π (π(π₯)) ππ₯ ππ₯ π π π (π(π₯) + π(π₯)) = π(π₯) + π(π₯) ππ₯ ππ₯ ππ₯ Product Rule π π (π β π)(π₯) = π(π₯) π(π₯) ππ₯ ππ₯ π + π(π₯) π(π₯) ππ₯ Quotient Rule π π π β² (π₯) β π(π₯) β πβ² (π₯) β π(π₯) ( ) (π₯) = ππ₯ π π2 (π₯) Composition Rule π (π°π)(π₯) = π β² (π(π₯)) β πβ²(π₯) ππ₯ π (π°π°β)(π₯) = π β² (π(β(π₯))) ππ₯ β πβ² (β(π₯)) β ββ²(π₯) U substitution composition π ππ(π’) ππ’ π(π₯) = β ππ₯ ππ’ ππ₯ π ππ(π£) ππ£(π’) ππ’ π(π₯) = β β ππ₯ ππ£ ππ’ ππ₯ Common Log Derivatives π 1 (ln(π₯)) = ππ₯ π₯ π 1 (log π (π₯)) = ππ₯ π₯ππ(π) Inverse Functions A function π(π₯) is one to one if for every pair of π₯1 , π₯2 then π(π₯1 ) β π(π₯2 ) (vertical line test) must pass vertical line test to have inverse (may have to restrict domain) Inverse function theorem: ππ₯ 1 Simply stated: ππ¦ = ππ¦βππ₯ Easy way to solve for an inverse is switch and solve (exchange all xβs and yβs, solve for y)