0lim ( ) = = indeterminate0x a

if f x then say

1 lim ( ) = undefined0x a

if f x then say

In any limit question, the first step is to replace x by the value given and check your answer as it may be: indeterminate, undefined, or any real numberCase.1: Indeterminate you should i lif I f ti t t f t

a very big numbera very small number

0simplify. In fractions you try to factor or

2lim .. 2.000..1x

has x approaching

2lim .. -1.999..9

xhas x approaching

lim .. -1.999..9has x approaching

y3

3

7 .7 .

take common denominator  then cancel like expressions. In radicals you conjugate or (rationalize). Case.2: Undefined you should apply the two sided does not imply 

differentiability

2lim .. 1.999..9

xhas x approaching

2lim .. -2.000..1

xhas x approaching

Case.2: Undefined you should apply the two sided limits for it may turn out  to be unequal and in this case you say the limit does not exist DNE Differentiablemeans 

continuous butcontinuity dtil

li ( ) li ( ) li ( ) ( )f f f f b

f(x) is a continuous function iff Rational functions: May admit Asymptotes Vertical (V) put q(x)=0 and find x=numberlim ( ) lim ( ) lim ( ) ( )

x bx b x bf x f x f x f b

If the relation is false the f(x) is a discontinuous function

Vertical (V) put q(x)=0 and find x=numberHorizontal (H) 

( )( )

p xyq x

( ) lim( )x

p xfindq x

Check your answer. If indeterminate then simplify. You get y=number, or as no (H)Rate of change=average rate of change=y/ x get y number, or  as no (H)ate o c a ge a e age a e o c a ge y/

=                  = dy/dx= slope of the tangent line to the curve = slope of the curve                   

at the point  of tangency  =                                              = f’(x) = y’=four‐step process

2 1

2 1

( )( )y yx x

0

( ) ( )limh

f x h f hh

Slope=y’= f’(x)=0  for a horizontal line=1/0 for a vertical line

Implicit derivative:  deriv. of y is y’ then  deriv of y2 is 2y y’ and deriv of 4y3 is

derivative of   y w.r.t  x  = velocity = V if f(x) stands for the position of a moving object at time x

h

Differentials:     y’=f’=dy/dx then dy=f’.dx    So if   Area=r2 then dA=2  r.dr  where  dA represents the change in  area and  dr the change in  r =increase or decrease of  r  if + ‐Related Rates: including the factor of time relating variables to timederivative of

deriv. of y is 2y.y  and deriv. of 4y is  12y2y’ while that of x has no x’ for  its value is 1. WHY? Because x is independent and y depends on xfor y=f(x). So in any implicit 

Related Rates:  including the factor of time  relating variables to time derivative of y  y ’ dy/dx because no more we are relating  y w.r.t x but in fact we are relating each w.r.t time ‘t’. So in this context derivative of y=change in rate of y=dy/dt. Same for dx/dtand deriv. of x2=2xdx/dt.             y also known as increment  in y.   same for  x= dx.     

While dy=df is the differential of y =part of the whole derivative y’ or f’.

equation you derivate with the presence of y’, then collect y’.

2 2 ' : (1. ' ) 2 ' 2 .

xy

xy

find y e y xe y conty x yy x

Function expression Derivative

Monomials, Binomials,and Polynomials

Y=3Y=xY=5x

Y’=0Y’=1Y’=5and Polynomials Y=5x

Y=kx , K in set RY=k.xnG(x)=k.un    for u=u(x) an expression in x

Y =5Y’=kY’=nk.xn‐1G’(x)=nk.un‐1 . u’

F(x)=5(3x4‐7x2+)11 F’= 55(3x4‐7x2+)10.(12x3‐14x)

Rationals (fractions) 11y xx

( )( )( )

u xg xv x

2

2

1'y xx

' ''( ) u v v ug x

Irrationals (Radicals) 1/2y x x

2( )g xv

1/21 1'2 2

y xx

k

Exponentials Y=ex     , f=k.eu(x) ,    h(x)=bx Y’=ex     , f’=k.eu(x).u’    ,  h’=bx .lnb    

/( ) . ( ) .[ ( )]n m nmg x k u x k u x

/ 1'( ) [ ( )] . 'm nkmg x u x un

( )

g=3e5x2‐ x g’=3e5x2‐ x .(10x‐1)             

Logarithms log.log b

y xy k x

01/ .ln1/ .ln

y xy k x b.log

log ( ).ln.ln ( )

b

b

y k xy u xy k xy k u x

/ .ln'/ .ln/. '( ) / ( )

y k x by u u by k xy k u x u x

Domain Range

*Lines (except special)     y=mx+b (‐, ) (‐, )

* Absolute lines            y=|x| (‐, ) [0, )

* d ( ) )*Quadratics (‐, ) [k,+)   if maximum(‐,k]   if minimum

*Rationals  q(x)≠0      Solve for x All y’s except horizontal Asymptote(s)

( )( )

p xyq x

2y ax bx c

*Exponentials                       a > 0 (‐, ) (0, )

*Logarithms (0, ),       f(x)>0, solve for x (‐, )

xy a

log ,y x logby x

lny x ln ( ) log ( )y f x or f x

*Radicals In odd (‐, )

In even [0, ) or  solve f(x)≥0

In odd (‐, )

In even (0, )

2 4

3 5

, ,..., ( )

, ,..., ( )

even

odd

x x f x

x x f x

Marginal cost, Revenue, Profit=C’(x), R’(x),P’(x) for Profit=R‐C and R=qp=quantity.price=xp given a price demand equation p.            Exact cost of producing (x+1)th item=C(x+1)‐C(x) which is estimated by C’(x)Break even points when R(x)=C(x) say (a,f(a)), (b,f(b))Profit= R(x)‐C(x) >0. If R > C then gain over Interval (a b) and if R < C then loss over two intervals

When asked to interpret  results, we say: “at a production level of …. the … is increasing + or decreasing ‐ at aInterval (a,b) and if R < C then loss over two intervals

(d1, , a)  (b, d2)  given (d1, d2) as domain of R and Cis  increasing +  or  decreasing   at a 

rate of ….per item”'( ) ( )( )'( )

f xRelative rate of change of f xf xf x

A price increase will decrease demand we say price and demandA price decrease will increase demand are directly proportionalDemand is inelastic: Price and revenue are again directly proportional

( ) 100%( )

'( ) ( ) ( )

f xPercentage rate of changef x

relative rate of change of demand f pElasticity of demand p E prelative rate of change of price f p

Demand is inelastic: Price and revenue are inversely proportionalA price increase decreases  revenue.   A price decrease increases  revenue