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## Exam for Wednesday

by: Brandon Short

21

0

3

# Exam for Wednesday Calculus 1151

Brandon Short
OSU

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This exam was supposed to be on the 17th. However, it was moved to the 12th. This should not be your only study tool, however, it is a good source.
COURSE
Math 1151 - Calculus 1 (11380)
PROF.
Professor Stephen Swihart
TYPE
Study Guide
PAGES
3
WORDS
KARMA
50 ?

## Popular in Calculus and Pre Calculus

This 3 page Study Guide was uploaded by Brandon Short on Monday October 10, 2016. The Study Guide belongs to Calculus 1151 at Ohio State University taught by Professor Stephen Swihart in Fall 2016. Since its upload, it has received 21 views. For similar materials see Math 1151 - Calculus 1 (11380) in Calculus and Pre Calculus at Ohio State University.

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Date Created: 10/10/16
Study Guide Exam Power of X:  d n (n­1) Power Rule:  /  Xdx= nX   d 2 Quotient Rule:  /  dxx)/g(x) = [(f(x) * g(x)) – (f(x) * g(x))] / [g(x)] Product Rule:  /  [f(x) * g(x)] = [f(x) * g’(x)] + [g(x) * f’(x)] dx Chain Rule:  /  dx(g(x))] = [f(g(x))]’ * g’(x)  Constant Multiple: Power rule d/dx(cf(x)) = c * f ’(x) Average and Marginal Cost Suppose C(x) gives the total cost to produce x units of a good cost. Sometimes,  C(x) = FC + VC * x FC = Fixed cost which does not change with units produced. VC = Variable cost which is the cost to produce each unit. C(x) = Average cost. C’(x) = Marginal cost, which is approximately the extra cost to produce one more unit beyond x  units. C’(x) = lim   C(x+∆x) – C(x) ∆x>0 ∆x Implicit Differentiation d d 4/3  1/3 Treat y as a function of x, so  /  (y(dx)4/3 = 4/3(y(x))1/3 * y’(x) =  /  y = 4/3y  *dx 1. Take d/dx of both sides. 2. Move all y’ terms to one side and non­y’ terms to the other.  3. Factor y’ out and divide to solve. 2 8 Terms like x y  require product rule as y is a function of x. Higher Order Derivatives 2 2 dy 1. To find d y/dx , first find  / .  dx dy dy 2. Differentiate both sides of  / . RHS dxll usually involve x,y, and y’ / dx  3. Replace y’ by what you know it to be. Memorize (sinx)’ = (cosx) (cosx)’ = (­sinx) 2 2 (tanx)’ = (sec x) (cotx)’ = (­csc x) (secx)’ = (secx + tanx) (cscx)’ = ­(cscx)(cotx) ­1 2 (sin x)’ = 1/√1 – x   ­1 < x < 1 ­1 2 (tan x)’ = 1/1 + x   All real x ­1 2 (sec x)’ = 1/│x│√x  + 1 x > 1, or x < ­1 ­1 2 (cos x)’ = ­1/√1 – x ­1 < x < 1 (cot x)’ = ­1/1 + x 2 All real x (csc x)’ = ­1/│x│√x  – 1 2 x > 1, or x < ­1 Logarithmic Differentiation This would be painful to differentiate with normal rules.  1. Take ln of both sides ln y =ln f(x). 2. Differentiate using implicit differentiation. Related Rates Procedure: 1. Identify known rates and desired rates. Label what you already know. 2. Find an equation relating the variables corresponding to known/desired rates, and only  these variables. 3. Take  / ,dtemembering implicit differentiation. 4. Solve for the desired rate and plug in known values.  Warning: Don’t plug in known values until the end. First Derivative Test For classifying critical points at local extrema. 1. Find f ‘(x) and factor completely. 2. Find the critical points of f, i.e. the places where f ‘ is 0 or DNE [but still in domain of f]. 3. Two parts: a. For critical points where f is continuous, draw a number line or sign chart for f ‘.  If f ‘ goes from positive to negative at a critical point, the critical point is a local  max. If f ‘ goes from negative to positive at critical point, the critical point is a  local minimum.  b. For a critical point where f is discontinuous, go back to the original definition of a local maximun/minimum.

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