DIFFERENTIATION
Laura Berglind
Avery Gibson
Definition of Derivative:
Lim
h0
f(x+h) – f(x)
h
Derivative= slope
Lim= limit
h0 = as h approaches zero
Notations for Derivative
•f’(x)
F prime
•Dy
dx
Derivative of y in
respect to x
•y’
y prime
Let’s Put the Definition to
Practice
 Example: f(x)=x²
 We know the answer is: f’(x)= 2x
Lim
h0
f(x+h) – f(x)
h
=
lim x²+2xh=h²-x²=
lim h(2x+h)
h0
h0
h
h
 When bottom H goes away, you can plug in 0 for h
Lim (2x+0) = 2x
h0
Thus, you can follow these
three simple steps!
Replace x with x+h and copy problem
2. Factor and get rid of h
3. Replace h with 0
1.
Let’s try one more!
f(x)= √x
Lim
h0
Think of this problem as: f(x)= x^1/2
f(x+h) – f(x)
h
=
lim (x+h)^1/2-x^1/2
h0
h
lim (x+h)^1/2-x^1/2
* (x+h)^1/2+x^1/2
h0
h
*multiple by the conjugate
(x+h)^1/2=x^1/2
lim
h0
x+h-x______
h [(x+h)^1/2 + x^1/2]
=
1 ______
(x+0)^1/2 +x^1/2
= 1____ =
2x^1/2
=
1/2x^-1/2
lim ___1_______
h0 (x+h)^1/2 + x^1/2
Before we move on…..
Old Rules to remember!
Lim sinh = 1
h0 h
Lim 1-cosh = 0
h0
h
*Sin(x+h) = sinxcosh + cosxsinh
*Cos(x+h) = cosxcosh - sinxsinh
Trigonometry
 f(x)=sinx
f’(x)=cosx
 Simple steps, we learned earlier:
Replace x with x+h and copy problem
2. Factor and get rid of h
3. Replace h with 0
1.
 Now we are going to use these steps to
prove f(x)=sinx
f’(x)=cosx
F(x)=sinx
Lim
h0
f(x+h) – f(x)
=
h
= Lim sinxcosh + cosxsinh- sinx
h0
Lim sin(x+h)-sinx
h0
h
=
Lim sinx(cosh-1) + Lim cosxsinh
h
= Lim sinx (0) + Lim cosx (1)
h0
f’(x)=cosx
h0
h0
=
h
Lim cosx
h0
h0
h
Now let’s prove that
f(x)=cosx
f’(x)=-sinx
Lim
f(x+h) – f(x)
h0
h
=
= Lim cosxcosh - sinxsinh- cosx
h0
h
=Lim cosx (0) – Lim sinx(1)
h0
h0
Lim cos(x+h)-cosx
h0
h
=
Lim cosx(cosh-1) - Lim sinxsinh
h0
= -sinx
h
h0
h
Now let’s prove that
f(x)=tanx
f’(x)=sec²x
*Think of f(x)=tanx as f(x)= sinx/cosx
Lim =
sin(x+h) – sinx
H0
cos(x+h) cosx
h
*now you need to find a common denominator and then flip!
Lim sin(x+h) cosx – sinxcos(x+h)
H0 _____cos (x+h) cosx_______
__h__
1
* _1_
h
Tangent continued…
Lim sin(x+h) cosx – sinxcos(x+h)
H0
h[cos(x+h) cosx]
Lim (sinxcosh + cosxsinh) * cosx- sinx (coscosh – sinxsinh)
H0
hcos(x+h) * cosx
Lim sinxcoshcosx + cos²xsinh – cosxcoshsinx + sin²xsinh *cancel out
H0
h cos (x+h) cosx
Tangent continued…
Lim
sinh (sin²x + cos²x)
h0
h cos(x+h) cosx
=
Lim
1 * 1_____
h0 cos(x+h)cosx
*side note: Never expand the bottom!!!
Lim
1______
h0 cos(x+0)cosx
*which equals sec²x!!!
=
Lim
h0
1___
cos²x
Memorize these trig
functions and their
derivatives!
 F(x)=sinx
F’(x)=cosx
 F(x)=cox
F’(x)=-sinx
 F(x)=tanx
F’(x)=sec²x
 F(x)=cotx
F’(x)=-csc²x
 F(x)=secx
F’(x)=secxtanx
 F(x)=cscx
F’(x)=-cscxcotx
*you can remember all the C’s have negative
derivatives!
SHORT CUTS!!!
1. Chain Rule
2. Product Rule
3. Quotient Rule
Chain Rule
 F(x)=u ^ n
F’(x)= nu ^ (n-1)
 The exponent goes out front and then
subtract 1 from the exponent
*used every time although may be embedded
in product or quotient
Example of Chain Rule
 f(x)= x³+6x
*bring the 3 and multiple it by the one is front of the x. then subtract 1
from 3 (the exponent).
*bring the 1 down to multiple it by the sixth. Then subtract 1 from the
exponent, which in this case is zero
 f’(x)=3x²+6xº
Thus, f’(x)=3x²+6
*side note: derivative of any constant is 0
For example: f(x)=5
f’(x)=0
Try Me!!!
f(x)=4x³+10x²-6x+100
And the answer is…
f(x)= 4x³+10x²-6x+100
f’(x)=12x²+20x-6+0
 Thus, f’(x)=12x²+20x-6
Product Rule
 Used when you are multiplying
 Equation: (F)(DS)+(S)(DF)
 F=first number or set of numbers
 DS=Derivative of second number
 S=second number or set of numbers
 DF=Derivative of first number
 For example: (x+2)(5x+2)
first second
Example of Product Rule
1. f(x)= (2x+1)^5 (5x-3)²
2. f’(x)=(2x+1)^5(2)(5x-3)(5) + (5x-
3)²(5)(2x+1)^4 (2)
3. F’(x)= 10(2x+1)^4(5x-3)[(2x+1)+(5x-3)]
4. F’(x)= 10(2x+1)^4 (5x-3) (7x-2)
Try Me!!!
f(x) = (4x+1)²(1-x)³
And the Answer is…
 f(x) = (4x+1)²(1-x)³
 f’(x)=(4x+1)²(3)(1-x)²(-1) + (1-x)³(2)(4x+1)(4)
 f’(x)= (4x+1)(1-x)² [-3(4x+1)+8(1-x)]
 f’(x)= (4x+1)(1-x)²[-12x-3+8-8x]
 f’(x)= (4x+1)(1-x)²[-20x+5]
 f’(x)= 5(4x+1)(1-x)²(-4x+1)
Quotient Rule
 Used when you are dividing.
 Equation: (B)(DT)-(T)(DB)
B²




B= the bottom number(the denominator)
DT= the derivative of the top number
T= the top number (the numerator)
DB= the derivative of the bottom number
 For example: 2+1x Top Number
5x-e Bottom Number
Example of Quotient Rule
 Y= (2-x)/(3x+1)
 Y’= (3x+1)(-1)-(2-x)(3)
(3x+1)²
 Y’=(-3x-1)=(-6=3x)
(3x+1)²
 Y’=___-7___
(3x+1)²
Try Me!!!
Y= 1+x²
1-x²
And the Answer is…
Y= 1+x²
1-x²
 Y’= (1-x²)(2x) – (1+x²)(-2x)
(1-x²)²
 Y’= 2x-2x³+2x+2x³
(1-x²)²
Y’= 4x___
(1-x²)²
Natural Log
 When taking the derivative of natural log: 1
over the angle * the derivative of the angle
 For example:
F(x)= ln(x²+3x)
F’(x)=
1__ *(2x+3)
x²+3x
Try Me Real Quick!
f(x)=ln (x³+5x²+6x)
And the Answer is…
 f(x)=ln (x³+5x²+6x)
 f’(x)=
1
(3x²+10x+6)
x³+5x²+6x
Derivative of e
 f(x)=e
f’(x)=e
 f(x)=e^2x
f’(x)=2e^2x
 f(x)=e^10x
f’(x)=10e^10x
When taking the derivative
of an exponent
*Copy function and then multiply by the
derivative of exponent
For example:
 F(x)=e ^sinx
 F’(x)=e^sinx (cosx)
ETA
 For only power of trig functions
 E= Exponent
 T= Trig
 A= Angle
 Copy the rest of problem!
For Example:
F(x)=sin^5(cosx)
F’(x)= (5)sin^4(cosx)*cos(cosx)*(-sinx)
Let’s try one more…
 F(x)= tan²x(5x²-6x+1)
 F’(x)=(2)tan(5x²-6x+1)*sec²(5x²-6x+1)*(10x-
6)
*copy rest of problem after derivative of exponent
(chain rule)
*copy problem after take derivative of trig
Let’s Try a REAl AP Multiple
Choice quesiton
 If f(x)=sin(e^-x), then f’(x)=?
a)-cos(e^-x)
b)cox(e^-x)+(e^-x)
c)cos(e^-x)-(e^-x) c)(e^-x) cos(e^-x)
d)-(e^-x)cos(e^-x)
If f(x)=sin(e^-x), then f’(x)=?
ANSWER:d)-(e^-x)cos(e^-x)
E1. sin^0(e^-x)  1
T2.cos(e^-x)
A3.(e^-x)(-1)
Let’s Review real quick
Try these!
 y=(4x+1)²(1-x)³
 Y=2-x
3x+1
 Y=3x^(2/3)-4x^(½)-2
y=(4x+1)²(1-x)³
 Y’= (4x+1)²(3)(1-x)²(-1)+(1-x)³(2)(4x+1)(4)
 Y’=-3(4x+1)²(1-x)²+8(1-x)³(4x+1)
 Y’=(4x+1)(1-x)²[-3(4x+1)+8(1-x)]
 Y’=(4x+1)(1-x)²[-12x-3+8-8x
 y’=(4x+1)(1-x)² 5(-4x+1)
Y=2-x
3x+1
 Y’=(3x+1)(-1)-(2-x)(3)
(3x+1)²
 Y’=-3x-1-6+3x
(3x+1)²
 y’= -7____
(3x+1)²
Y=3x^(2/3)-4x^(½)-2
 Y’=(2/3)(3)(x^-1/3)(1)-(4)(1/2)(x^-1/2)(1)-0
 Y’=2x^-1/3-2x^-1/2
 Y’=2(x^-1/3 – x^-1/2)
Implicit Differentiation
 Used when x and y are in the problem and
they can’t be separated.
 When taking the derivative of y. . . You still do
it in respect to x. . . Everything is in terms of x
 Dy
dx always there when taking derivative of
y
*solve for dy/dx
Let’s Try one
F(x)= X²+y²=25
1. 2x+2y(dy/dx) = 0 *take derivative like
normal, except whenever you take the
derivative of y, you must add in dy/dx
2. *Now, solve for dy/dx
2y(dy/dx)=-2x
3. dy= -2x
= -x_
dx
2y
y
Your Turn!
 F(x)=(2x+1)^5+3y²=6
F(x)=(2x+1)^5+3y²=6
 F’(x)=(5)(2x+4)^4(2)+6y(dy/dx)=0
 6y(dy/dx)=-10(2x+4)^4
 dy= -10(2x+1)^4
dx
6y
 dy= -5(2x+1)^4
dx
3y
 *side note: if the dy/dx were to cancel out,
that means that the derivative does not exist!
Logarithmic Differentiation
 Done when there is a variable in the exponent
 3 rules to remember 
Multiplication  addition
2. Division  subtraction
3. Exponents  multipliers
1.
For example: y=2^x
 Lny= xln2 *RULE: when you add the ln, the

1
y


exponent comes down in front as so.
*now take the derivative of both sides!
dy = x(0)+ln2(1)
dx
Dy/dx=ln2(y)*plug in your original equation for y
Dy/dx=ln2(2^x)
Formula:
*y
= a^x
Dy/dx = a^x lna
(this would help with the first bullet on the
previous slide)
Now your turn
 F(x)=x^sinx
F(x)=x^sinx
 Lny=sinxlnx
 1 dy = sinx 1 (1) + lnxcosx
x dx
x
 Dy/dx= x^sinx[sinx/x + lnxcosx]
^we mult. By y!
Almost done. . .
Now let’s try an FRQ
 1975 AB2
 This FRQ is about particle motion, which involves
position, velocity and acceleration.
 In order to solve these kind of problems you must
be able to take the derivative!
 For the derivative of position=velocity
 And the derivative of velocity=acceleration
1975 AB 2
 Given the function f defined by f(x)=ln(x2-9).




A. Describe the symmetry of the graph of f.
B. Find the domain of f.
C. Find all values of x such that f(x)=O.
D. Write a formula for f-1(x), the inverse function of f,
for x >3.
THE END
© Laura Berglind and Avery Gibson, 2011, AP Calculus AB, Autrey