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Week of September 19th

by: Brandon Short

Week of September 19th Calculus 1151

Brandon Short

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Buy as needed. These are great if you missed the week or just don't feel like taking notes!
Math 1151 - Calculus 1 (11380)
Professor Stephen Swihart
Class Notes
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This 4 page Class Notes was uploaded by Brandon Short on Monday October 10, 2016. The Class Notes belongs to Calculus 1151 at Ohio State University taught by Professor Stephen Swihart in Fall 2016. Since its upload, it has received 4 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
 Calculus notes for week of 9/19/16 3.6 Derivatives as Rates of Change Velocity is measured as: V  ave(t+∆t) or s(b) – s(a)           ∆t   b – a (Change in position over change in time.) S’’(t) = V’(t) = A(t) (From left to right: S=Position, V= Velocity, and A=Acceleration) 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 3.7 Chain Rule How do we differentiate a composition of functions f(g(x)) or (f*g)(x)? Chain Rule: If g is differentiable at x and f is differentiable at g(x), then (f*g) is differentiable.  [f(g(x))]’ = f ‘(g(x)) * g’(x)  Pseudo Proof [f(g(x))]’ = lim  f(g(x+h)) – f(g(x))        h>0 h     = lim  f(g(x+h)) – f(g(x)) * g(x+h) – g(x)        h>0     g(x+h) – g­(x)          h Since g is differentiable, it is continuous, so u = g(x+h) > g(x) as h>0, Then, = lim      f(u) – f(g(x)) * lim  g(x+h) – g(x)   u>g(x)    u­g(x)   h>0         h Chain Rule Alternative Version If y is a differentiable function of u, and u is a differentiable function of x, then y(u(x)) is a  differentiable function of x, with dy = dy * du dx    du    dx 3.8 Implicit Differentiation We know how to differentiate many explicitly defined functions f(x), How can we differentiate implicitly defined functions such as F(x,y) = x – x y  + e  = 8? 2 3 y Implicit Differentiation Treat y as a function of x, so  /  (y(dx)4/3 = 4/3(y(x))1/3 * y’(x) =  /  y = 4/3y  *dx 4/3  1/3 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 1. To find d y/dx , first find  / . dy 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. 3.9 Derivatives of Logarithms and Exponentials  We know  /  e dx ex x d What is  /  dxx? x lnx Since e  and lnx are inverse functions, e  = x for x>0 Differentiate both sides.  d/  e  =  /  x dx dx lnx e  * (lnx)’ = 1 > x * (lnx)’ = 1 (lnx)’ = 1/x for all x = 0. It’s good to know  /  ln│x│ = 1/x for all (x<0) dx Logarithms with other Bases d What is  /  dxg │x│b Apply change of base formula: Log Ab= log A  a log argument    logaB log aase Use a = e  log b = lnA    lnB d /dxlog │b│= d/dx (ln│x│)(lnB) = (1/lnB) (1/x) = (1/xlnB) Exponentials other than e. How do we differentiate b  or f(x) ? g(x) Trick: Convert to base e first. d x d lnb x xlnb /dxb  =  / dxe )  = d/dx e = e xln * 1 * lnb = b  * lnb Logarithmic Differentiation F(x) =  (2x      +      (      –    ) x^3  5 8 4 (1 + x + x ) 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. a. 1/y * y’ =  /  ln f(x) dx 3. Solve for y’. y’ = y  /  ln fdx) Ln y = ln(2x2 + 1)5 + ln(3e  ­ x )  – ln 4  – ln(1 + x – x ) 5 8 2 ­2x 3 3 5 = 5ln(2x  + 1) + 2ln(3e  – x ) – x ln4 – 8ln(1 + x – x ) 2 5 4 = ­3x ln4 – 8 * [1/(1 + x – x )] *(1 – 5x ) 3.10 Derivatives of Inverse Trig Functions (sin x)’ = 1/√1 – x   2 ­1 < x < 1 (tan x)’ = 1/1 + x   2 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


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