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MOLLY LEDFORD
SYDNEY ENTEN
AP CALCULUS 1ST PERIOD
TABLE OF CONTENTS
Definition of the Derivative
Limits
Chain Rule
Product Rule
Quotient Rule
Implicit
Logarithmic
ETA
Second Derivative
DEFINTITION OF THE
DERIVATIVE
The derivative of f is the function whose value at x is
the limit
f’(x)= lim f(x+h) – f(x)
h0
h
provided this limit exists.
APPLICATION OF LIMITS
f(x) = 2x^2 + 5 x
f’(x)= 4x+ 5
lim f(x) = 2(x+h)^2 + 5(x+h) – 2x^2 + 5x
h0
h
lim 2(x^2 +2xh + h^2)+5x+5h-2x^2+5x
h0
h
lim 2x^2+4xh+2h^2+5x+5h-2x^2+5x
h0
h
lim h(4x+2h+5)
h0
h
lim 4x +2h+5
h0
4x+2(0)+5= 4x+5
THEOREMS FOR LIMITS
lim sinh= 1 lim 1-cosh= 0
h
h
h0
h0
PROVE THE DERIVATIVE OF SIN(X)
USING THE DEFINITION OF THE
DERIVATIVE
f(x)= sinx
f’(x)= cosx
lim sin(x+h)-sinx
Use angle addition formula
h0
sin(x+h)=sinxcosh+cosxsinh
h
cos(x+h)=cosxcosh-sinxsinh
lim sinxcosh+cosxsinh-sinx
h0
h
lim sinx(cosh-1) + lim cosxsinh
h0
h
h
lim sinx(0) + lim cosx
Theorum usage here:
h0
f’(x)=cosx
(cosh-1)/h becomes 0
and (sinh)/h becomes 1
**THINGS TO REMEMBER**
f(x)= sinx
f(x)=cosx
f(x)=tanx
f(x)=secx
f(x)=cotx
f(x)=cscx
f’(x)=cosx
f’(x)= -sinx
f’(x)=sec^2x
f’(x)=secxtanx
f’(x)=csc^2x
f’(x)=cscxcotx
All of these can be proved by using the definition
of the derivative and theorems for limits as well
as precalculus prerequisites.
WELCOME TO THE CHAIN
RULE
f(x)= ax^n
f’(x)= anx^(n-1)
Every time you take a derivative, a chain rule is
required. IT IS ALWAYS THERE.
REMEMBER: the derivative of a constant equals zero.
CHAIN RULE IN ACTION
EXAMPLE ONE
f(x)= x^2
f’(x)=2x
EXAMPLE TWO
f(x)= 5x^6 + 7x^-3
f’(x)= 30x^5 – 21x^-4
TRY ME
f(x)= 7x^4 – 3x^3
f’(x)= 28x^3- 9x^2
f(x)= (8x-3)^2
f’(x)= 2(8x-3)(8)
MEET THE PRODUCT RULE
f(x)g’(x)+ g(x)f’(x)
FDS + SDF
First times Derivative of Second
+
Second times Derivative of First
EXAMPLE
f(x)= (3x+1)^2*(5x^2-1)^3
f’(x)= (3x+1)^2*3(5x^2-1)^2(10x) + (5x^2-1)^3(3x+1)(3)
BEFRIEND THE PRODUCT
RULE
EXAMPLE
f(x)=(4x+1)^2*(1-3x^2)^4
f’(x)= (4x+1)^2*4(1-3x^2)^3(-6x)+ (13x^2)^4*2(4x+1)(4)
SIMPLIFICATION IS APPRECIATED BUT NOT
REQUIRED. BEWARE OF CARELESS MISTAKES
WHEN SIMPLIFYING.
TRY ME
f(x)= (2x-1)^3(3x-2)
f’(x)= (2x-1)^3(3) + (3x-2)(3)(2x-1)^2(2)
HELLO QUOTIENT RULE
This is the quotient rule:
BDT-TDB
B^2
Bottom times Derivative of
Top minus Top times
Derivative of Bottom over
Bottom squared
EXAMPLE
f(x)= (3x-2)^4
(5x-1)^2
f’(x)= (5x-1)^2(4)(3x-2)^3(3)-(3x-2)^4(5x-1)(5)
(5x-1)^4
GETTING TO KNOW
QUOTIENT RULE
EXAMPLE
f(x)= (2x-3)^2
(8x+6)^4
f’(x)= (8x+6)^4(2)(2x-3)(2)-(2x-3)^2(4)(8x+6)^3(8)
(8x+6)^8
TRY ME
f(x)= (5x-2)
(x^2+1)
f’(x)= (x^2+1)(5)-(5x-2)(2x)
(x^2+1)^2
f’(x)= (5x^2+5)-(10x^2-4x)
(x^2+1)^2
f’(x)= -5x^2+4x+5
(x^2+1)^2
SALUTATIONS FROM IMPLICIT
What is implicit differentiation?
It is the process of finding the derivative of a
dependent variable in an implicit function by
differentiating each variable separately by
expressing the derivative of the dependent
variable as a symbol and then solving the
resulting expression for the symbol.
dy
dx = derivative of y with respect to x
EXPLORING IMPLICIT
EXAMPLE
f(x)=x^2+y^2=7
2x+2y dy
dx= 0
y dy
dx = -x
dy = -x
dx y
APPLYING IMPLICIT
EXAMPLE
f(x)=x^2+xy+3y^(2) = 4
dy
dy
2x+x dx +y+6y dx = 0
dy
dx (x+6y) = -2x – y
dy = -2x – y
dx x+6y
TRY ME
f(x)=3x^2 + 6xy + y^2 = 6
=6x + 6x dy
+y(6) + 2y
dx
dy
(6x
+
2y)
=
-6x
–
6y
dx
dy = -6x – 6y
dx 6x+2y
dy = -3x – 3y
dx 3x+y
dy
dx
=0
LOVE ME SOME LOGARITHMIC
DIFFERENTIATION
Logarithmic differentiation relies on the chain rule. In
order to solve, you must also remember the properties
of logarithms.
REMEMBER:
-multiplication becomes addition
-division becomes subtraction
-exponents become multipliers
y=a^x
=a^xlna
ln(1)=0
lnex=x
ln(x/y)= lnx-lny
lne=1
ln(xy)= ln(x)+ln(y)
HOW TO SOLVE LOGARITHMIC
1. First take the logarithms on both sides. We
get, log y = x log x.
2.Differentiating on both sides with respect to x.
3. Apply the chain rule if necessary and solve
further.
4. Multiply both sides by y to get the final
solution.
LOGARITHMS AND YOU
EXAMPLE
y= 3^x
lny=xln3
1 over angle = derivative of angle therefore
1 dy = x(0)+ln3
y dx
dy = 3^xln3
dx
LIFE WITH LOGARITHMS
EXAMPLE
y=x^sinx
lny= sinxlnx
1 dy = sinx(1/x)+ lnxcosx
Y dx
dy = x^sinx[(sinx)/x +lnxcosx]
dx
TRY ME
y= x^cos3x
lny=cos3xlnx
1 dy = -3(sin3x)lnx + 1 cos3x
Y dx
x
dy = [-3(sin3x)lnx + cos3x/x]y
dx
= [-3(sin3x)lnx+cos3x/x]x^cos3x
ETA
EXPONENTS-bring exponent in front and copy
TRIG-take derivative of trig and copy what’s inside
parenthesis
ANGLE-take derivative of inside parenthesis
EXAMPLE
d (tan^2(sin3x)) E 2tan(sin3x)
dx
T sec^2(sin3x)
A sec^2(sin3x)(cos3x)(3)
RULES TO REMEMBER
d e^x = e^x
dx
d e^3x=e^3x
dx
d e^sinx= e^sinx(cosx)
dx
ETA AND LOGARITHMIC
EXAMPLES
y=(x^3sec5x)^1/2
lny= ½(x^3sec5x)
= 1/2(ln(x^3)+ ln(sec5x))
= ½(3lnx+ln(sec5x))
1 dy = ½[ (3/x) + (1/sec5x)(sec5x)(tan5x)(5)
y dx
dy = (3/2x +5tan5x)y
dx
dy = (3/2x+5tan5x)(x^3sec5x)^1/2
dx
CONTINUED
y=sin^5(cosx)
y’= (5sin^4x(cosx))(cos(cosx)(-sinx))
DERIVATIVE OF e
d/dx e^u = e^ u (du/dx)
Copy the function given and take the derivative of
the angle.
DERIVATIVE OF Natural Log
1/angle times the derivative of the angle
y=ln u
y’= (1/u)(du/dx)
1977 AB 7 BC 6
Let f be the real-valued function defined by f(x)
= sin3(x) + sin3|x|.
a. Find fʹ(x) for x > 0.
b. Find f’(x) for x < 0.
c. Determine whether f(x) is continuous at x =
0. Justify your answer.
d. Determine whether the derivative of f(x)
exists at x = 0. Justify your answer.
A) For x > 0
F(x) = sin3x + sin3x = 2sin3x
F’(x) = 6sin2xcosx
B) For x > 0
f(x) = sin3x + sin3(-x) = sin3x - sin3x = 0
F’(x) = 0
ANSWER
C) f(0) = 0
Lim+ f(x) = lim +2sin3x = 0
X ->0
x->0
Lim- f(x) = lim+ 0 = 0
X ->0
x->0
Since Lim f(x) = 0 = f(0), the function of f is continous at x = 0
X ->0
D) F’(x) = lim f(x+h) – f(x) if the limit exists
H -> 0
h
At x = 0
Lim f(h) - f(0) = 0
H -> 0
h
Therefore, lim f(h) - f(0) and so f’(0) exists and equals 0.
H -> 0
h
SECOND DERIVATIVE
The second derivative follows the same steps as
the first derivative, only applied on the first
derivative.
EXAMPLE
y= 20x^4
y’= 80x^3
y”=240x^2
Bibliography
http://www.analyzemath.com/calculus/DefDerivati
ve/DefDerivative.html
http://archives.math.utk.edu/visual.calculus/3/logd
iff.1/index.html
http://www.tutorvista.com/content/math/calculus/
differentiation/logarithmic-differentiation.php
http://sites.google.com/site/autreyap/calculus
All of our notes, quizzes, and tests on
differentiation.
©
Molly Ledford
Sydney Enten
2011