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Cheat Sheet Definition and fx  h fx  If yfx   then the is defined to be fx lim . h0 h

If yfx  then all of the following are If yfx all of the following are equivalent equivalent for the derivative. notations for derivative evaluated at x  a . df dy d df dy fx y   fx  Dfx fa y   Dfa   xa  dx dx dx dxxa dx xa

Interpretation of the Derivative If yfx  then, 2. fa is the instantaneous of 1. mfa  is the of the change of fx at x  a . line to yfx  at x  a and the 3. If fx is the of an object at of the tangent line at x  a is x then fa is the of given by yfafaxa   . the object at x  a .

Basic Properties and Formulas If fx and g x are differentiable functions (the derivative exists), c and n are any real , d 1. cf  cf  x 5. c  0 dx  2. fg  fxgx    d nn1 6. xnx – Rule dx   3. fg f g fg – d 7. fgx  f gxgx  ffgfg  dx 4.  2 – This is the gg

Common Derivatives d d d x 1 cscxxx csc cot aaaxx ln  dx dx dx d d d sinxx cos cotxx csc2 eexx dx dx dx d d 1 d 1 cosxx sin sin1 x  lnxx , 0 dx dx 1 x2 dx x d 2 d 1 d 1 tanxx sec cos1 x  lnxx , 0 dx dx 1 x2 dx x d d 1 d 1 secxxx sec tan 1 loga xx , 0 dx tan x  2 dx xln a dx1 x

Visit http://tutorial.math.lamar.edu for a complete of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Chain Rule Variants The chain rule applied to some specific functions. d nn1 d 1. fx  nfx  f  x 5. cosfx  f x sin  fx  dx   dx   d d 2. eefx  fx fx 6. tanfx  f x sec2  fx  dx  dx   d fx d 3. ln fx  7. secfx() f () x sec fx () tan fx () dx f x dx fx d d 1   8. tan fx  2 4. sinfx  f x cos  fx  dx  dx 1fx 

Higher Order Derivatives The is denoted as The nth Derivative is denoted as df2 dfn fx  f2  x and is defined as fxn  and is defined as dx2 dxn

 nn1  fx  fx  , i.e. the derivative of the fx  f x, i.e. the derivative of first derivative, fx. the (n-1)st derivative, fxn1 .

Implicit Differentiation Find y if e29xy xy 32 sin y  11 x. Remember yyx  here, so products/quotients of x and y will use the product/quotient rule and derivatives of y will use the chain rule. The “trick” is to differentiate as normal and every time you differentiate a y you tack on a y (from the chain rule). After differentiating solve for y.

e29xy 29yxyxyyyy 3 22  2 3  cos   11 11 2e29xy 3xy 22 29ee29xyyxyxyyyyy 29 xy 32cos11 22 3  29xy329e xy cos y 29xy329eexy cos1123 y y  2922 xy xy

Increasing/Decreasing – Concave Up/Concave Down Critical Points x  c is a critical of fx provided either Concave Up/Concave Down 1. If fx  0 for all x in an I then 1. fc 0 or 2. fc doesn’t exist.  fx is concave up on the interval I. Increasing/Decreasing 2. If fx  0 for all x in an interval I then 1. If fx  0 for all x in an interval I then  fx is concave down on the interval I. fx is increasing on the interval I. 2. If fx 0 for all x in an interval I then Inflection Points x  c is a of fx if the fx is decreasing on the interval I.  concavity changes at x  c . 3. If fx 0 for all x in an interval I then fx is on the interval I.

Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Extrema Absolute Extrema Relative (local) Extrema 1. x  c is an absolute maximum of fx 1. x  c is a relative (or local) maximum of if fc fx for all x in the domain. fx if fc fx  for all x near c. 2. x  c is a relative (or local) minimum of 2. x  c is an absolute minimum of fx fx if fc fx  for all x near c. if fc fxfor all x in the domain.   1st Fermat’s If x  c is a critical point of fx then x  c is If fx has a relative (or local) extrema at 1. a rel. max. of fx if fx 0 to the left x  c , then x  c is a critical point of fx. of x  c and fx 0 to the right of x  c .

Extreme Value Theorem 2. a rel. min. of fx if fx 0 to the left If fx is continuous on the closed interval of x  c and fx 0to the right of x  c . ab, then there exist numbers c and d so that, 3. not a relative extrema of fx if fx is 1. acdb, , 2. fc is the abs. max. in the same on both sides of x  c .

ab, , 3. fd is the abs. min. in ab, .    2nd Derivative Test If x  c is a critical point of fx such that Finding Absolute Extrema To find the absolute extrema of the continuous fc 0 then x  c fx on the interval ab, use the 1. is a relative maximum of fx if fc  0 . following process. 2. is a relative minimum of fx if fc  0 . 1. Find all critical points of fx in ab, . 3. may be a relative maximum, relative 2. Evaluate fx at all points found in Step 1. minimum, or neither if fc  0 . 3. Evaluate fa and fb. 4. Identify the abs. max. (largest function Finding Relative Extrema and/or value) and the abs. min.(smallest function Classify Critical Points value) from the evaluations in Steps 2 & 3. 1. Find all critical points of fx. 2. Use the 1st derivative test or the 2nd derivative test on each critical point.

Mean Value Theorem If fx is continuous on the closed interval ab, and differentiable on the open interval ab, fb fa  then there is a acb such that fc . ba

Newton’s Method

th st fxn If xn is the n guess for the root/solution of fx 0 then (n+1) guess is xxnn1  fxn provided fxn exists.

Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Sketch picture and identify known/unknown . Write down equation relating quantities and differentiate with respect to t using implicit differentiation (i.e. add on a derivative every time you differentiate a function of t). Plug in known quantities and solve for the unknown . Ex. A 15 foot ladder is resting against a wall. Ex. Two people are 50 ft apart when one The bottom is initially 10 ft away and is being starts walking north. The angle changes at 1 pushed towards the wall at 4 ft/sec. How fast 0.01 rad/min. At what rate is the is the top moving after 12 sec? between them changing when   0.5 rad?

We have    0.01 rad/min. and want to find x is negative because x is decreasing. Using x . We can use various trig fcns but easiest is, and differentiating, xx xy2215 2  2 xxyy  2  0 sec sec tan   50 50 1 After 12 sec we have x 10 124  7 and We know  0.5 so plug in   and solve. 22 x so y 15 7 176 . Plug in and solve sec 0.5 tan 0.5 0.01    for y. 50 7 x  0.3112 ft/min 717601 yy   ft/sec Remember to have in ! 4 4 176

Optimization Sketch picture if needed, write down equation to be optimized and constraint. Solve constraint for one of the two variables and plug into first equation. Find critical points of equation in range of variables and verify that they are min/max as needed. Ex. We’re enclosing a rectangular with Ex. Determine point(s) on yx2 1 that are 500 ft of fence material and one side of the closest to (0,2). field is a building. Determine dimensions that will maximize the enclosed .

2 22 Minimize fd x02 y and the Maximize Axy subject to constraint of constraint is yx2 1. Solve constraint for xy2 500. Solve constraint for x and plug x2 and plug into the function. into area. xy2212 fxy    2 Ay500 2 y  xy500 2 2 2 500yy 2 2 yy12  yy  33  Differentiate and find critical point(s). Differentiate and find critical point(s). 3 Ayy 500 4 125 fy 23  y 2 By 2nd deriv. test this is a rel. max. and so is By the 2nd derivative test this is a rel. min. and the answer we’re after. Finally, find x. so all we need to do is find x value(s). xx2 3 1 11   x 500 2 125  250 22 2 The dimensions are then 250 x 125. The 2 points are then 1 ,3 and  1 , 3 . 2 2 2 2

Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins