Filetype PDF | Posted on 25 Jan 2023 | 2 years ago
Authentication
digital resources
199x Filetype PDF File size 0.04 MB Source: www.craftonhills.edu
Calculus Cheat Sheet Calculus Cheat Sheet
Derivatives Chain Rule Variants
Definition and Notation The chain rule applied to some specific functions.
nn−1
fx+−hfx d d
( ) ( ) ′ ′
1. fx=nfxfx 5. cosfx=−fxsinfx
() () () () () ()
′ ( ) ( )
If y= fx then the derivative is defined to be fx=lim .
( ) ()h→0 h dx dx
d fxfx d
() ′ () ′ 2
2. ee=fx 6. tanfx= fxsec fx
() () () ()
( ) ( )
If y=fx then all of the following are If y= fxall of the following are equivalent dx dx
( ) ( ) ′ d
fx
equivalent notations for the derivative. notations for derivative evaluated at xa= . d ( ) ′
7. secf(x)= f(x)secf(x)tan fx()
3. ln fx = ( [ ]) [ ] [ ]
()
( )
dx
dxfx
dfdyd dfdy ()
′′
fx=y====fxDfx
() () () ′′
( ) fa=y===Dfa
() () ′
fx
dxdxdx xa= dxdx d d −1 ( )
8. tan fx =
x==axa ()
′ ( ) 2
4. sinfx= fxcosfx
() () ()
( )
dx 1+fx
dx ()
Interpretation of the Derivative
′ Higher Order Derivatives
If y=fx then, 2. fa is the instantaneous rate of
( ) ( ) th
′ The Second Derivative is denoted as The n Derivative is denoted as
1. m=fa is the slope of the tangent change of fx at xa= .
( ) ( ) 2 n
2 df n df
′′ () ()
fx==fx and is defined as fx= and is defined as
line to y=fx at xa= and the 3. If fx is the position of an object at () () 2 () n
( ) ( ) dx dx
equation of the tangent line at xa= is ′
time x then fa is the velocity of
( ) ′ ′
nn−1
′′′ () ()
fx=fx, i.e. the derivative of the
() ()
( ) fx=fx, i.e. the derivative of
() ()
′ ( )
given by y=fa+−faxa.
( ) ( )( ) the object at xa= .
′ n−1
first derivative, fx. st ( )
( ) the (n-1) derivative, fx.
( )
Basic Properties and Formulas
If fx and gx are differentiable functions (the derivative exists), c and n are any real numbers, Implicit Differentiation
( ) ( )
′ 2xy−932
Find y if e+xy=+sinyx11 . Remembery=yx here, so products/quotients of x and y
′ ′ d ( ) ( )
1. cf=cfx
( ) () 5. c=0
dx() will use the product/quotient rule and derivatives of y will use the chain rule. The “trick” is to
′ ′′ d differentiate as normal and every time you differentiate a y you tack on a y′ (from the chain rule).
2. f±g=±fxgx
( ) () () nn−1
6. x=nx – Power Rule
dx( ) After differentiating solve for y′.
′ ′′
3. fg=+fgfg – Product Rule
( ) d
′′
7. fgx=fgxgx 2xy−9223
() () ()
( ) ( )
( ) ′′′
e2−9y+3xy+2xyy=+cosyy11
′ dx ( ) ( )
′′
ffg−fg 2xy−922
4. = – Quotient Rule This is the Chain Rule 2x−−9y2xy9223 11−−23exy
2 ′′′′
2ee−9y+3xy+2xyy=cosyyy+11 ⇒=
gg () 329xy−
2xyy−−9ecos
32x−−9y2xy922 ()
′
2xy−9ee−cosyy=11−−23xy
()
Common Derivatives ( )
d d d
xx Increasing/Decreasing – Concave Up/Concave Down
x=1 cscx=−cscxxcot a=aaln
() ( ) ()
dx dx dx() Critical Points
d d d
2 xx xc= is a critical point of fx provided either Concave Up/Concave Down
sinxx=cos cotxx=−csc ee= ( )
dx( ) dx( ) dx( ) ′′
1. If fx> 0 for all x in an interval I then
′ ′ ( )
1. fc=0 or 2. fc doesn’t exist.
d d 1 d 1 ( ) ( )
cosxx=−sin sin−1x = lnxx=>,0 fx is concave up on the interval I.
( ) () ( )
( ) 2 ( )
dx dx 1−x dxx Increasing/Decreasing ′′
2. If fx< 0 for all x in an interval I then
d 2 d 1 d 1 ( )
tanxx=sec −1 lnxx=≠,0
( ) () ′
cos x =− 1. If fx>0 for all x in an interval I then
dx ( ) 2 dxx ( ) fx is concave down on the interval I.
dx 1−x ( )
d d 1 fx is increasing on the interval I.
secx=secxxtan d 1 logxx=>,0 ( )
( ) −1 ()
tan x = (a )
dx ( ) 2 dxxaln ′ Inflection Points
2. If fx<0 for all x in an interval I then
dxx1+ ( )
xc= is a inflection point of fx if the
fx is decreasing on the interval I. ( )
( ) concavity changes at xc= .
′
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 Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins
Calculus Cheat Sheet Calculus Cheat Sheet
Extrema Related Rates
Absolute Extrema Relative (local) Extrema Sketch picture and identify known/unknown quantities. Write down equation relating quantities
1. xc= is an absolute maximum of fx 1. xc= is a relative (or local) maximum of 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.
fx if fc≥fx for all x near c.
if fc≥fxfor all x in the domain. ( ) ( ) ( )
( ) ( ) 2. xc= is a relative (or local) minimum of Ex. A 15 foot ladder is resting against a wall. Ex. Two people are 50 ft apart when one
2. xc= is an absolute minimum of fx The bottom is initially 10 ft away and is being
( ) fx if fc≤fx for all x near c. starts walking north. The angleθ changes at
( ) ( ) ( ) pushed towards the wall at 1 ft/sec. How fast 0.01 rad/min. At what rate is the distance
if fc≤fxfor all x in the domain. 4
( ) ( ) is the top moving after 12 sec? between them changing when θ = 0.5 rad?
st
Fermat’s Theorem 1 Derivative Test
If xc= is a critical point of fx then xc= is
If fx has a relative (or local) extrema at ( )
( ) ′
1. a rel. max. of fx if fx>0 to the left
xc= , then xc= is a critical point of fx. ( ) ( )
( ) ′
of xc= and fx<0 to the right of xc= .
( ) ′
′ x′ is negative because x is decreasing. Using We have θ =0.01 rad/min. and want to find
2. a rel. min. of fx if fx<0 to the left
Extreme Value Theorem ( ) ( ) x′ . We can use various trig fcns but easiest is,
If fx is continuous on the closed interval ′ Pythagorean Theorem and differentiating, ′
( ) of xc= and fx>0to the right of xc= . xx
( ) 222 ′
′′ secθ=⇒=secθtanθθ
ab, then there exist numbers c and d so that, ′ x+y=15⇒2xx+=20yy 5050
[ ] 3. not a relative extrema of fx if fx is
( ) ( ) 1
After 12 sec we have x=10−=127and
1. a≤≤c,db, 2. fc is the abs. max. in the same sign on both sides of xc= . ( 4 ) We knowθ =0.5 so plug in θ′ and solve.
( ) 22 x′
so y =15−=7176. Plug in and solve sec0.5tan0.50.01 =
ab, , 3. fd is the abs. min. in ab, . ( ) ( )( )
[ ] ( ) [ ] 2nd Derivative Test for y′. 50
If xc= is a critical point of fx such that x′ = 0.3112 ft/sec
( ) 1 7
′′
Finding Absolute Extrema 7−+176yy=0⇒= ft/sec Remember to have calculator in radians!
′ ( 4 ) 4176
fc=0 then xc=
To find the absolute extrema of the continuous ( )
function fx on the interval ab, use the ′′
( ) [ ] 1. is a relative maximum of fx if fc< 0.
( ) ( ) Optimization
following process. ′′
2. is a relative minimum of fx if fc> 0.
1. Find all critical points of fx in ab, . ( ) ( ) Sketch picture if needed, write down equation to be optimized and constraint. Solve constraint for
( ) [ ] 3. may be a relative maximum, relative one of the two variables and plug into first equation. Find critical points of equation in range of
2. Evaluate fx at all points found in Step 1. ′′ variables and verify that they are min/max as needed.
( ) minimum, or neither if fc= 0.
( )
Ex. We’re enclosing a rectangular field with 2
3. Evaluate fa and fb. Ex. Determine point(s) on yx=+1 that are
( ) ( ) 500 ft of fence material and one side of the closest to (0,2).
4. Identify the abs. max. (largest function Finding Relative Extrema and/or field is a building. Determine dimensions that
value) and the abs. min.(smallest function Classify Critical Points will maximize the enclosed area.
1. Find all critical points of fx.
value) from the evaluations in Steps 2 & 3. ( )
st nd
2. Use the 1 derivative test or the 2
derivative test on each critical point. 222
Minimize f=d=xy−02+− and the
( ) ( )
Maximize A= xy subject to constraint of 2
Mean Value Theorem xy+=2500. Solve constraint for x and plug constraint is yx=+1. Solve constraint for
If fx is continuous on the closed interval ab, and differentiable on the open interval ab, 2
( ) [ ] ( ) into area. x and plug into the function.
2
22
fb−fa x=y−12⇒f=xy+−
( ) ( ) A=−yy5002 ( )
′ ( )
then there is a number a<
no reviews yet
Please Login to review.
