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Evaluating limits to chain rule derivatives

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Study note

Constance Martin

United States of America

DEFAULT (US)

AP Calculus AB/BC

Units 1 & 2: Limits, Discontinuity, Theorems, & Derivatives

Differentiation: definition and basic derivative rules

Applying the power rule

1-1 nater of change
* remember ROC is just another word for slope *
f(x) = 3x+1
the hoc of
F(x) = 3.
RoC fora linear
function is a
Constant value
*
Two types of ROC
-two endpoints;
• Average-
finding average slope
We want to find the
RoC from
average
X= 8 +0 X=2
f(2)=3
f(0) = 0-1
F(2)-F(0) = 3-D
2-0.
F(x)=x²-1
the Rol of
F(x) =
*For all other non-linear
equations the hoc
varies. *
f(0) = -1 m = 2
F(2)=3.
Instantaneous-uses. 1 point to find slope
exactly that instant
X2-1
|||||
|||||
8114122
IZ
The seant line of a function (equation) connects two points on
its graph.
y=mx+b
3=2(2) + b
3= 4+ b
-4-1 = b
secant line equation
y=2x-n
Secantline.
*The slope of the secant line is the avg. RoC blt those pts..
F(-1) - F(3) = 1-29 --28
3
·4
What is the avg. BoC OF F(x) = x³ +2
on the interval E1,3]?
F(-1) = 1
f(3) = 29
We can write the secant line equation using the slope + a point
F(x)=x²-1
↓
=
(7.) * write the secant line equation for F(x) = x³+2 over [-1,3].
F(x) = x3+2
y=mx+b
F(-1) = 1
F(3)=29
m=7
F(x)=x2-1
how can we determine
the instantaneous
RoC at x = 1?.
y=7x+08
inst.
Roc
↑
||||||||
tan line
•V=mx+b
0=2(1)+b
0=2+b
-2=0
y=2x-2
* To find the tan line algebraically, we can
Choose points on the graph arbitrarily. close.
to the x-value of the tan line *
Find the inst. Roc OF F(X) = x3-2x² +5X-4@ x = 0
m=5
Y == 4
+1=7(-1) + b
•1=-7+6
Ⓡ = b
8.
--4=5(0)+b
O+b
- 4= b
[V=5x-4
Find the inst RoC. off(x) = x² = a @ x = -3
X+3.
8|m=1]
AP Classroom - Co questions.
m = 2
(1,0) 1-2:
Evaluating
Limits:
Graphically + Tabularly.
A limit is a value that a function/table/graph approaches..
While we CAN have an exact value defined for our
limit it is not required we can just APPROACH it
We can write limit notation as: as
lim
limit X
The value for the limit is the y-value that corresponds to
the X-value we approach.
F(x)=x²-1
g(x) = x+1
h(x) =
X-> a for a given function - The font approaches
・gets
limf(x) = -1
x70
limf(x)=3
X-72
X-2
Cyo
lim h(x)=und,
x->2
F(x) = x² + 5x+6₂
X2-9
lim (@geg(x) = 2
lim g(x) = 1
limg(x) = und.
8117
limg(x) =
- 1/2
If our limit yields. 0/0 for direct
Substitution, we say that limitis in
the indeterminate form -
Must be solved algebraically
or graphically. At
limf(x) = 1/6
X-7-3
in order for the limit to
exist, I must approach
the same y-value both
below
above the given
x-value
*A function is defined at
a
given x-value wherever
the circle is filled,
A HHHH
Timg(x) = 2
X-7-4
limg(x)= und.
X-7-1.
limg(x) = -1
X-72 ·1-3: Evaluating Limits Algebraically.
lim 7²²-9 = 6₂
X-73
X-3
Factoring, we can factor
lim X2-9
X-73 X-3
Jim x2+3x=10 = (x+5
= 1
lim
lim X+5
X->2 X-1.
it Ifa limit is evaluated in the indeterminante
form it can be solved using a table!
graph or it can be solved using
algebraic manipulation
the numerator + denominator,
lim5 x2-3x-10-
X-75
6x+5
lim
X-75 X-1
lim
X-74
fim
(x+3)(x-3) -> lim X+3 = 6
X-3 X-3
Xr3
X+2 =
3+√√x+q
x
驳
3-√2+ 3+√√x+9
3+√019
규
x-303-ind. form
(x+5)(x-2)
Rationalization: used when radical expressions occur
in numerator or denominator
(5)(x+2)
to rationalize, we multiply by the conjugate
of the radical expression
do NOT oversimplify *.
-5)(x-1)
8122123
2-4 (√x+2)(x-4)(√x+2) (X-4)(√x+2)
√X-2 (√x+2)
x +2+x-2√x-4
X-4
(X)(3+√x+9)
9 + 3√x¹9 −3+x+a}−(X+q)
3+3 = 6/1
=
lim √x+2=4
X-74
XX)(3+√x + 9)
9-(x+9)
9-x-의 3 common denoms. Of complex fractions!
lim.
X->1
1 - 1x
X-1.
lim
X-70 X+2
x
(1-4) (+) ²
2
(+= 1/X 1/²
X = 1 = X-1
Of
Separate into two fractions
using multiplication
veciprocal
yet a comm. denom, for
Fractional part.
OUV
(袋) (衣)文
lim 12 = 1
2-2 (X)
2 = (x+2)
2x74
2x+4 1-4: Continuity
that
Wecansay a graph is continuous if it goes on.
w/O stopping stopping over a given interval
it. if you can draw the
wlo,
・graph.
picking up your pencil then it is
Continuous
F(x),
We are discontinuous
@X=0 therefore f(x) is
NOT Continuous
There are 3 types of discontinuities.
Intinite: Occurs when graph approaches ±; vertical.
Basymptotes occur nere
8128123
Removal (hole): occurs when there is I instant where the
graph is NOT defined.
3. Jump: Occurs when there is a gap between x-values;
only occurs in piecewise functions.
Discontinuities commonly occur where we divide
·by. 0. if a factor exists in both the num. + denom.,
its Ois a removable discontinuityt.
If a factor only exists in the denom., its. O is.
an infinite discontinuity*
F(X) is.
discontinuous @ X= 1 + X = -1
F(x) = X-1
X2-1.
X-1
(X+1.)(X-1).
X=1 is a removable discontinuity ble it
OCCurs in both the num. + denom..
X=-1 is an infinite discontinuity blc
it only occurs in the denom. F(X) =
1
"X-4 x≤4
XQ-160 x>4.
§.
This graph is continuous.
A left-handed limitis the value a graph approaches
from the lefthand side 1-∞
Tim F(X).
X-> a
A right-handed limit is the value a graph approaches
from the righthand sid. 1
(F(x) =
For a graph to be continuous, it must meet the following
Ⓒlim f(x) .= lim f(x)
X->a
X-7a+
x-a+
(2) F(a) must exist
3 lim F(X)
X-7a-
F(x)=
=
X2-9 X23.
lim G(x) = f(a)
X-7a+
lim f(x) = 5
X-73-
MUST MEMORIZE
on apiecewise func., we want to
check where the X-value function.
ruce changes. A
·lim f(x) = 0
x-73.+
X+4K
for what value of Kis
F(X) Continuous?
1-5: Limit
When evalutating.
infinity, we conside
becomes arbit
lim 4x-3
X-700
lim
X->∞
3x²+5x-1
4x2-5x+6
2x²+3x+1
lim
role of
when com
the denomi
NUM DE
2 DENOMY
3 EQUALE
a.) lim.
Бха
X-700 8x24
X-18
c.) lim 5x³+7x²-
2x-9 1-5: Limits Approaching Infinity
When evalutating the lim of a function as. X approaches
infinity, we consider. what happens to the y-value as x
becomes arbitrarily large () or arbitrarily small (-00)
=.0
`lim
X-7
4x-3.
3x² +5x-1
lim bx212
X-
lim 4x-3 = 0
X->- 3x²+5x-1
5x2+2
limo
4x2-5x+2=2
2x²² +3x+1
rife of dominance
lim
X->-∞
a:) lim 5x²-12 5/8
+ 4x +1_
c.) lim 5x³+7x2-4x²
X->∞
2x-9
lim
X->-∞0
4x2-5x+2=2
2x2+3x+1
when comparing the degree of the numerator to
the denominator.....
NUM > DENOM: num dominates denom. ; ±∞
2 DENOM > NUM: denom, dominates num. ; [0]
3 EQUAL: co dominant ; lim = coefficient of num, over
Coeff. Of denom..
b.) lim.
X-100
916123
3 x-q
2x² +7x4
d.) lim √9²-4
xo x
315 1-6: Theorems About
Limits
The Squeeze Theorem (AKA the Dinching Theorem.
that for
Functions such that f(x) ≤g (x) ≤ h(x), if lim F(x) =
lim h(x) = 4, then the lim y(x) =L
X->a
x->α.
x->a
F(x) = x²
9.(x) = xxx 3
= = x²
squeezed
Since x3 is in the middle
of x² + x2 since lim x2 = lim -x2₁
X-70
X-70
GYO
V
then lim x³ > 0.
X-70
9/11
Theorem (IVT) states
The Intermediate value the
that if a function is continuous, and asb, then there
exists a value c such that f(a) ≤ f(c) ≤ f(b)
++++||||.
t
The IVT states that if (-3, 4) +
(2,-5) are coordinates on F(x)
there is a value for all y- valjes.
between F5,4] 2-1: Derivatives + the Power Rule
The word derivative is interchangeable w/ the word slope
Differentiation is the process of calculating a derivative
y = a →> the derivative of yea is written as dy
・Liebnitz
Notation
dx
F(x)= a -> The derivative of f(x) is written as f'(x)=. THE
I prime o
of X
Lagrange
Notation
F(x) = x²
C
(X, F(X))
(X+h, f(x+h))
m = f(x+h)-f(x)
x+h-x
Tim (x+h)2-x2
X-70
• Tim 4xh+262 +5h.
h-₂0
Can be interpreted as "the
derivative of y wl respect to X".
lim.
não f(x+) - F(x)
n
lim 2xh th ² → K (2x+h) lim 2xth = 2x
>>
=
X2+2xh+n2-x2
9/25/23
Find an equ, for the deriv. of f(x) when F(x) = 2x² + 5x-1
Ilim F(x+h)-f(x) 2(x+h)2 +5 (x+h)-1-(2x2+5x-1)
haso n
2(x2+2xh+h²) + 5x + 5h-1-2x2-5x-1-2x²+4xh+24² +5x+5
h
mux+an+5) = lim 4x+2h+5 = 4x+5
n-0
2x2² The Power Rule can be used to calculate the deriv. of any polynomial fure
- IF F(x) = ax", then f(x)=n·a·x^-!
F(X) = (x³-1
F'(X) = 18X2
F(X) = 3x²
F(x) = 3·4x*.
F(x) = 12x "-1
F(x) = 12x3
"F"(x) = 12 (2) ³
12 (8)
96
f(2)=3(2)"
3(16).
48.
F(x)=2x²
·8x3
3 F(x) = -7x9
-63x8
to differentiate.
O Multiply exponent by Coefficient.
Subtract 1 from exponent
Find deriv.
Slope @x=2
Write tan line equ.@x=2
m = 96
(2,48)
Y=Y₁ = m(x= x₁).
Y-48=96(x-2)
for #4, write tan line equ: @ X = 2
2) F(x) = 1 17x³
51x2
(4) f(x) = 3x³
F'(x) = 9x²
y-24-36(x-2)
m=36
(2,24) The derivative of a constant is = 0
3 = 3x therefore derivative = 0
This extends to symbols used in place of numbers
Such as Tande.
*If a function is multiplied by a scalar such as 5f (x)
f(x) = 3x² + 8 x - 1
5. F(X) = 1²0x² + 40x-5
the derivative is 5x'(x)
F'(X) = 6x + 8
X5F'(x) = 30x +40
*If a function is written as the sum I difference of component *
Functions, its derivative is the sum of the derivatives of those
Component terms.
h(x) = 3x² - 6x +1
h'(x) = 6x-6.
g(x) = x3 +2x2-4x+1 · g'(x) = 3x² + 4x -4
= +
f(x) = x²³ +2×²+3x2-4x-6x+1+1
F(x)=x²³ +5x²-10× +2
F'(x)=3x2+10X-10
f'(x) = h'(x) + g'(x)
F'(x) = 3x2+4x+6x-4-6
F'(x)= 3x2 +10X-10 2-2 Derivatives Using the
Product Rule
We use the product rule to evaluate the derivative of two
quantities being multiplied together.
f(x) = g(x);h(x)
('(x) = g(x) · n(x) + h'(x) + g(x) denir.
f(x) = (2x-1)(3x2+5x).
F'(x) = 2(3x² + 5x) + (6x+5) (2x-1)
(x) = 6x² +10x + 12x2 - 6x +10x-5
F(x) = 18 2²2² +14x-5
f(x) = (2x-1) (3x²+5x)
F(x) = 6 x ²³ +10x²-3x²-S
f(x) = (x³ +7x 2-5x
F'(x) = 18x²+14x-5
g(x) = 5x³ sinx
·g.'(x) = = (Oxsinx + cosx(5x2)
n(x):JX (SX-3)
n(x) = x"2.5x-3
h'(x) =
"The deriv. of the first
term times the second
plus the second term times
the
--112
n'(x) = "¹2x¹¹² (5 x-3) + 5 (x¹2)
(5x3)+5x2
h'(x) = 2x¹2
2√x
(5x-3)+5X
XS+ E-X9 = (X), 4
2√x
h'(x) = (5x-3)√x + 55x
2 x
not
necessary F(x) = tan (x)
F(x) = tan(x) · X-4
f'(x) = secax-x-4
F'(x) = Secax - 4 tanx.
хи
x5
-4x* tanx
2-3: Quo
The quotic
the deriva
h(x).
h'(x
F(x) =
F(x)=
F'(x
F
дех 2-3: Quotient Rule
The quotient rule is a used when we need to take
the derivative of a division problem / Fraction
h(x) = f(x)
g(x)
h'(x) = g(x). F'(x) = f(x) · g'(x)
•·ho.
h(x) = _hi.
✓h(x) = Hod Hi-HidHo
Ho Ho
F(x) = 5x² -> F'(x) = x² 20x³ -
ха
(x²)2
F'(x) = 20x5
хи
F(x) = 5x² = 5x²
x2
F'(x) = 10x
10x5
this is a
- Correct answer.
Write the tan line equation for g(x) @ X=1
5x² - 2x
= 10x5=10x
F(x) = x² + 8x +2 -> F'(x) = (x³ +6x2_4x)· (2x+8) (x2 + 8×+2)(3x²³ +12×)
x3+6х2-4х
(x3+6x2-4x) 2
g(x) = 5x2-4x+1 -> 1g|(x)= (3x2+2)· (10x-4)= (5x2-4x+1)+(6x)
g'(1) = (5) (16) = (²02) (6) = 30 - 12 = 18
25
25
· g(1) = 5-4+1 = ² (1₁²) m = 18/25
5.
y-2/5=¹/25 (x-1) F(x) = secax
f(x) = tanx ->
r(x) = sinx − r(x) =(cosx) (Cosx)(sinx)(sinx)
CO SX
Cosax
f'(x) = Cos²x + sin 2x
C052x
F'(X) =
Cosax
f(x) = 3x²= 5x +1 find tan live equ. @ x = -2
x2+4.
12 + 10 + 1
F₁(x) = (x2+4) (6x-5)-(3x2-5x+1)(2x)
(x2+4)2
|f|(-2) = (8) (-17) -(23) (-4) = -1/16.
F. (2) = 12+20+1 = 23
V- 23/8 = - 11/16 (x+2)
= бесах
y=2318
'm = = "/16
di
dy
(f(x)
F(x
F(U)
F'(U)
F(x)=
· y = (5x ³ - 6x
_Y = 04.
・dy=4j³d
dy = 4(5x
dy (60x² 2-4 Chain Rule
Chain Rule is used to differentiate quantities that
Can not be broken down
*. If it's in parentheses, use chain rulet
n(x) = f(g(x))
•h'(x) = F
F! ( g ( x ) ) ) : g' ( x )
y = (x²+4) ² || U = x2+U
. du = 2x
y= u²
dy=audo]
dy=2(x²+4) (2x)
dy=4x²² +16x
F(x) = (3x²-4) ¹2
F(U) =
0112
F'(U) = 120 ¹/2 du.
U= 3x2-4
du = 6x
F(x) = 1/2 (3x²2-45¹/2² (6x) = 3x (3x2-4) ¹ / 2 = 3x
53x2-4
U= 5x³-x +3
du= 15x2-6
10128 123)
y=(5x³-6x+3) 4
Y = 04
・dy = 4u ³ du
•dy = 4(5x³-6₂+3) ³ (15x²-6);
dy (60 x ²²-24) (5x ³-6x+3) ³
set whatever is in parentheses
equal to U
Differentiate U
Substitute U into original.
equation
Differentiate original.
equation
Sub. everything back.
in
Simplify algebraicly.
mult..
Choice y = {(²=EXICO)
y = eu
dy e du
dy = ex² + 5x + 6 (2x + 5).
f(x) = sin(2x²)
F(u) = sin u
· F.'(U) = cos(u) du
F'(x) = cos(2x²)-4x
F'(x) = 4x Cos(2x2)
F(x) = 1n (4x² - 8x+3)
F(U) = In (u)
· F'(U) = — du
(f'(X) =
4х2-8х+3
f'(x) = 8x-8
4x2-8x+3
U= x²+x+6
du=2x+5.
h(x) = f(g³(x))
n(0) = f(u)
U= 2x2
du = 4x
(8x-8).
U= 4x2-8x +3
du = 8x -8.
J = g(x)
h' (u) = f'(U) dj
h'(x) = f(g(x))· (g₁x))
Write tan line equ. @X=4 of f(x) = √√.X²-7
f(x)=(x²-7) 2
U=X2-71
du= ax
f(u) = 01/2
F'(U) = 1/20¹2 du
f'(x) = 1/2(x²-1)-¹1/12 (2x),
(x²-7) - 1/2 (x) 3-1- implicit
is the method of taking the derivative;
of an equation w/ multiple variables.
Apply the standard rules (power, product,
quotient + chain) but we included.
derivative
we take
the derivative of a nonmain variable.
+ y² = 25
2x+2ydy=0
ta.
10₁2 -2x = x2
204d4-2=
2xdx = -2ydy.
-2y
differentatio
2
X²dy = 2x4
ха
VADS
= 2xdx = dy
-2y -2ydx.
dx
dy = 2x+2 = x+1
・201. .Joy.
Write the fan, line equation @ (2,1))
dx = -2xy = -24
ха
ах
2x=dx = = *
Fay.
Solve for dx using algebra
dy (2₁1) = -2(1) = -1
10131/23
20µdy = 2x+2
201
201
x²x=4
2xy + x2dy=0
(1-1=1(x-2)
-2xy. Write the fan line equ: @ (1,2).
- xdy = -8x+y=-2
dx = -8x+y=2
ax
4x2-XX+2x=4
8x-y-xdy+2=0
-8x +Y.

Units 1 & 2: Limits, Discontinuity, Theorems, & Derivatives

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Units 1 & 2: Limits, Discontinuity, Theorems, & Derivatives

Can't find what you're looking for? Explore other subjects.

Knowunity is the # 1 ranked education app in five European countries

Knowunity was a featured story by Apple and has consistently topped the app store charts within the education category in Germany, Italy, Poland, Switzerland and United Kingdom. Join Knowunity today and help millions of students around the world.

4.9+

13 M

#1

950 K+

Still not sure? Look at what your fellow peers are saying...

I love this app so much [...] I recommend Knowunity to everyone!!! I went from a C to an A with it :D

The application is very simple and well designed. So far I have found what I was looking for :D

Love this App ❤️, I use it basically all the time whenever I'm studying