Calculus Cheat Sheet Derivatives Definition and Notation fx h fx If yfx then the derivative is defined to be fx lim . h0 h
If yfx then all of the following are If yfx all of the following are equivalent equivalent notations 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 rate of 1. mfa is the slope of the tangent change of fx at x a . line to yfx at x a and the 3. If fx is the position of an object at equation of the tangent line at x a is time x then fa is the velocity 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 numbers, d 1. cf cf x 5. c 0 dx 2. fg fxgx d nn1 6. xnx – Power Rule dx 3. fg f g fg – Product Rule d 7. fgx f gxgx ffgfg dx 4. 2 – Quotient Rule This is the Chain Rule 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 sin1 x lnxx , 0 dx dx 1 x2 dx x d 2 d 1 d 1 tanxx sec cos1 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 set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Chain Rule Variants The chain rule applied to some specific functions. d nn1 d 1. fx nfx f x 5. cosfx f x sin fx dx dx d d 2. eefx fx fx 6. tanfx f x sec2 fx dx dx d fx d 3. ln fx 7. secfx() f () x sec fx () tan fx () dx f x dx fx d d 1 8. tan fx 2 4. sinfx f x cos fx dx dx 1fx
Higher Order Derivatives The Second Derivative is denoted as The nth Derivative is denoted as df2 dfn fx f2 x and is defined as fxn and is defined as dx2 dxn
nn1 fx fx , i.e. the derivative of the fx f x, i.e. the derivative of first derivative, fx. the (n-1)st derivative, fxn1 .
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 29yxyxyyyy 3 22 2 3 cos 11 11 2e29xy 3xy 22 29ee29xyyxyxyyyyy 29 xy 32cos11 22 3 29xy329e xy cos y 29xy329eexy cos1123 y y 2922 xy xy
Increasing/Decreasing – Concave Up/Concave Down Critical Points x c is a critical point of fx provided either Concave Up/Concave Down 1. If fx 0 for all x in an interval 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 inflection point 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 constant 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 Derivative Test Fermat’s Theorem 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 sign 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 function 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 number acb such that fc . ba
th st fxn If xn is the n guess for the root/solution of fx 0 then (n+1) guess is xxnn1 fxn provided fxn exists.
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Related Rates Sketch picture and identify known/unknown quantities. 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 quantity. 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 distance 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, Pythagorean Theorem and differentiating, xx xy2215 2 2 xxyy 2 0 sec sec tan 50 50 1 After 12 sec we have x 10 124 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 717601 yy ft/sec Remember to have calculator in radians! 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 field with Ex. Determine point(s) on yx2 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 area.
2 22 Minimize fd x02 y and the Maximize Axy subject to constraint of constraint is yx2 1. Solve constraint for xy2 500. Solve constraint for x and plug x2 and plug into the function. into area. xy2212 fxy 2 Ay500 2 y xy500 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