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Unformatted text preview: MATH 135 (U of T) Study Sheet
and if g ′(a ) is not 0, then Review
LINEAR QUADRATIC EQUATION:
2
for ax 2 + bx + c = 0 , x = − b ± b − 4ac
2a
LOGARITHMS:
log b ( x ⋅ y ) = log b ( x ) + log b ( y ) ,
⎛ x⎞
log b ⎜ ⎟ = log b ( x) − log b ( y ) ,
⎜ y⎟
⎝⎠ lim
x →a x→ 3 EXAMPLE:
Evaluate 3 1/3 + log 2 2 3 = log 8 8 1 /3 + log 2 2 3 = 1
10
+3=
3
3 EXPONENT LAWS: b m ⋅ b n = b m+ n , ⎛ b ⎞ = b m−n ,
⎜ n⎟
⎜⎟
m (b m ) n = b m⋅n , ⎝b ⎠
1
b = b , b0 = 1 h →0 ln 3 + 2ln 5 = eln 3 ⋅ e 2 ln 5 = eln 3 ⋅ eln 25 = 3 ⋅ 25 = 75 TRIGONOMETRY: sin 2 u + cos2 u = 1, 1 + tan 2 u = sec 2 u ,
1 + cot 2 u = csc2 u ,
sin (u ± v ) =
cos (u ± v ) = cos u cos v m sin u sin v , sin u cos v ± cos u sin v , tan (u ± v ) = tan u ± tan v
1 m tan u tan v Limits and Continuity
The function f(x) has limit L, as x approaches a,
denoted lim f x = L if given any ε > 0 ,
x →a the derivative of both the numerator and the
denominator.
3
2 lim x + 3 x − 10 x
3 x − 3x + 2 x
d3
x + 3 x 2 − 10 x
dx
= lim
=
x→2 d
x 3 − 3x 2 + 2 x
dx
x→2 ( ) ( () ) 2 −5+ f (x + h ) − f (x ) .
h EXAMPLE:
Find the derivative of the function
2
f (x ) = (x + 1) by first principles. SOLUTION: f (x + h ) − f (x )
h
(x + h + 1)2 − (x + 1)2
= lim
h →0
h
f ′( x ) = lim = lim (x 2 h→0 )( ) + 2 xh + 2 x + h 2 + 2h + 1 − x 2 + 2 x + 1
h 2 xh + h 2 + 2h
h(2 x + h + 2 )
= lim
h →0
h→0
h
h
= lim 2 x + h + 2 = 2(x + 1) = lim h →0 TECHNIQUES FOR FINDING LIMITS:
First try substitution, factoring
If you’re taking the limit as x → ∞ of a
quotient, and substitution yields an invalid
answer, then try first to divide both the
numerator and the denominator of the limit
expression by its highest power of x
If the limit expression contains absolute
values, then try breaking the limit up into two
onesided limits
If the limit expression is a quotient, and if
factoring doesn’t work, then try l’Hopital’s Rule RULES FOR DIFFERENTIATION:
For functions u,v and for constants c, n: () dy
du ,
cu n = cnu n −1
dx
dx
dy
dv
du ,
(uv ) = u + v
dx
dx
dx
du
dv
v
−u
dy ⎛ u ⎞
dx
dx
⎜ ⎟=
dx ⎝ v ⎠
v2
DERIVATIVES TO REMEMBER: EXAMPLE: x2 − 5 − 1+ x lim there exists a δ > 0 such that f(x)  L < ε for
all x satisfying 0 < x – a< δ. Evaluate lim[ f (x ) ± g ( x )] = lim f ( x ) ± lim g ( x ) , SOLUTION: Let L be the limit. lim[ f ( x ) ⋅ g ( x )] = lim f ( x ) ⋅ lim g ( x ) , ⎛ x 2 − 5 − 1 + x ⎞⎛ x 2 − 5 + 1 + x ⎞
⎜
⎟⎜
⎟
⎠⎝
⎠
L = lim ⎝
⎛ x2 − 5 + 1 + x ⎞
2
x →3
x −9⎜
⎟
⎝
⎠ x2 − 9 x →3 d
1 du d u
du
,
,
(ln u) = ⋅
(e ) = e u
du
u dx dx
dx
du
du d
,
(a ) = a u (ln a )
(sin x ) = cos x ,
dx
dx dx
d
d
(cos x ) = − sin x , (tan x ) = sec 2 x ,
dx
dx
d
2
(cot x ) = − csc x ,
dx More free study sheet and practice tests at: x→a x→a x→a x→a x→a ⎡ f ( x ) ⎤ lim f ( x ) ,
= x→a
lim ⎢
x→a g (x ) ⎥
⎦ lim g ( x )
⎣
x→a [ n x→a x→a lim[ f ( x )] = lim f ( x )
x→a x 1+x⎞
⎟
⎠
5
=
24
1+x⎞
⎟
⎠ h →0 2 lim ln 3 + 2ln 5 SOLUTION:
e 0 form. Therefore take
0 3x 2 + 6 x − 10 14
=
=7
x→2 3x 2 − 6 x + 2
2 EXAMPLE:
Simplify e f ′(x ) = lim x − 3x + 2 x Note that this limit is in SOLUTION: + 3 )⎛
⎜
⎝ (x + 2 )(x − 3 )
− 3 )⎛ x 2 − 5 +
⎜
⎝
(x + 2 ) DIFFERENTIATION BY FIRST PRINCIPLES: 2 EXAMPLE: )(x The Derivative x 3 + 3x 2 − 10 x . lim +3 (x = lim SOLUTION: Evaluate log8 2 + log2 8. (x x→ 3 f ( x)
f ′( x )
= lim
x → a g ′( x )
g ( x) x →2 log b ( x n ) = n ⋅ log b ( x) log 8 8 = lim n ( = lim x→3 L’HOPITAL’S RULE:
If the limit of the quotient of differentiable functions f ( x ) and g ( x ) are of types 0 or ∞ ,
0
∞ = lim x→3 . ) x2 − 5 − 1 − x
x2 − 9 ⎛ x2 − 5 + 1 + x ⎞
⎜
⎟
⎝
⎠ (
( )
) x2 − x − 6 x − 9 ⎛ x2 − 5 + 1 + x ⎞
⎜
⎟
⎝
⎠
2 d
(sec x ) = sec x ⋅ tan x ,
dx
d
(csc x ) = − csc x ⋅ cot x
dx
CHAIN RULE: f ( x ) = (a o b )( x ) , then
f ′( x ) = a ′(b( x )) ⋅ b ′( x ) If More free Study Sheets and Practice Tests at: www.prep101.com More free study sheets and practice tests at
Step 1. Take the ‘ln’ of both sides (to find an
expression of the form ln y = ln[ f ( x )] ). EXAMPLE:
Evaluate the derivative of the function
y = tan sec e sin( x ) . (( )) SOLUTION: (( )) y ′ = sec 2 sec e sin ( x ) ⋅ (( d
sec e sin ( x )
dx )) [ ( (( Step 2. Simplify )) )] )( )) [ ( )] )( y ′ = sec 2 sec e sin ( x ) ⋅ sec e sin ( x ) tan e sin ( x ) ⋅ e sin ( x ) (( y ′ = sec sec e
2 sin( x ) ))⋅ [sec(e sin( x ) )tan(e ln ( f ( x )) by using the sin( x ) )]⋅ [e ( d sin ( x )
e
dx d
(sin (x ))
dx sin( x ) cos( x) ) 1 dy d
⋅
= (ln f ( x )) ).
y dx dx
Step 4. Solve for dy dx . (thus: Step 5. Express the answer in terms of x only
(substitution f(x) for y). EXAMPLE:
Find the derivative of the function y = x x . Take the natural logarithm of both d u(x)
du
c = c u ( x ) ⋅ ln c ⋅
dx
dx sides: ln Find the derivative of the function y = x ln x . Differentiate both sides: EXAMPLE: x
1 dy 1 −1 / 2
⎛ ln x ⎞
=x
= x −1 / 2 ⎜1 +
ln x +
⎟
y dx 2
x
2⎠
⎝ y = 7x SOLUTION: dy x ⎛ ln x ⎞
Solve for dy :
=
⎜1 +
⎟
dx dx
2⎠
x⎝
x y ′ = 7 x ln 7 ⋅ d
(x )
dx y ′ = 7 x ln 7 Applications of the Derivative HIGHER ORDER DERIVATIVES: Critical points: The values of x ∈ domain of f(x)
such that f ′ x = 0 or f ′( x) is not defined. f ′′( x) = d ⎛ dy ⎞ .
⎜⎟ () dx ⎝ dx ⎠ f ′( p ) = 0 TECHNIQUES OF DIFFERENTIATION First derivative test: If Implicit differentiation: to differentiate an implicit
function
Step 1. Take the derivative of both sides with
respect to x. Use the Chain Rule on terms
involving y (and note that the derivative of y with changes from negative to positive at p, then f
has a relative minimum at p. If f ’ changes from
positive to negative at p, then f has a relative
maximum at p.
Second derivative test: If f ′ p = 0 and dy dx .)
Step 2. Collect all terms involving dy dx f ′′( p ) > 0 , then f respect to x must be left as on one side of the equation.
Step 3. Solve for dy dx . EXAMPLE: 2 2 Differentiate the function x + xy + y = 7. SOLUTION: ⇒ ⇒ Recall that by definition, a curve is concave up
when its second derivative is positive. We find: dy
2x
=
dx 3 + x 2 and f ’ () dx 2 = and 2( 3 + x 2 ) − 4 x 2
(3 + x 2 ) 2 = 6 − 2x 2
>0
(3 + x 2 )2 (when  3 < x < 3 )
CURVE SKETCHING
Step 1. Find the intercepts.
Step 2. Find all the asymptotes.
Step 3. Find critical points and the intervals of
increase/decrease.
Step 4. Find inflection points and the intervals in
which the function is concave up/down. EXAMPLE:
Given a function with the property ekx (2kx 2 − 1) ,
x2
2 f ′( x) = k k > 0 , determine the intervals on which f is increasing and
decreasing and the location of any local
maximum and minimum values of f. SOLUTION:
We find that e kx (2kx 2 − 1)
= 0 when x = ±1 / 2k ,
x2
2 f '( x) = k hence these points are the locations of any local
maximum or minimum values of f. Now we
examine the change of the sign of the derivative
of f at these points. has a relative minimum at p. () If f ′( p ) = 0 and f ′′ p < 0 , then f has a
relative maximum at p. If f ′( p ) = 0 and f ′′( p ) = 0 , then the test fails.
f ′′( p ) > 0 on an interval, then f(x)
is concave up on that interval; if f ′′( p ) < 0 , then Concavity: If x −1/
More free study sheet ′′and practice tests=at: 2k Differentiate both sides: 2x + y + x ⋅ 2 Where is the curve y = ln (3 + x ) concave up? d2y SOLUTION: POWER RULE: EXAMPLE:
SOLUTION: properties of logarithms.
Step 3. Differentiate both sides with respect to x y ′ = sec 2 sec e sin ( x ) ⋅ sec e sin ( x ) tan e sin ( x ) ⋅ (( www.prep101.com dy
dy
+ 2y ⋅
=0
dx
dx dy
(x + 2 y ) = −2 x − y
dx dy
2x + y .
=−
dx
x + 2y LOGARITHMIC DIFFERENTIATION:
to differentiate a function with complicated
exponent/a product of several functions, etc. It’s
also used to differentiate functions that have an x
in both the base and in the exponent. f(x) is concave down on that interval. A point of
inflection occurs when f ( p ) changes sign (and
thus concavity).
Vertical asymptote: The line x = a is a vertical
asymptote for the graph of the function f(x) if
and only if lim f ( x ) = ±∞ or
x→a + lim f ( x) = ±∞ . x→a − Horizontal asymptote: The line x = b is a
horizontal asymptote for the graph of the
function f(x) if and only if lim f ( x) = b or
x → +∞ lim f ( x) = b x → −∞ We conclude that and x = 1 / 2k correspond to a local maximum
and a local minimum for f respectively. OPTIMIZATION PROBLEMS
Step 1. Determine what we’re trying to
maximize/minimize and write an equation for
this.
Step 2. Write a second equation from additional
information given in the problem, isolate one of
the two variables, and substitute this into the first
equation.
Step 3. Take the derivative of this equation, set it
to zero and solve for the remaining variable.
Step 4. Plug the value of this variable into the
original equations to solve for the remaining
variables. Helping students since 1999 ...
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 Math, Calculus

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