differentiation laura berglind avery gibson. definition of derivative: lim f(x+h) – f(x) h 0 h...

DIFFERENTIATION

Laura BerglindAvery Gibson

Definition of Derivative: Lim f(x+h)

– f(x)h0 h

Derivative= slopeLim= limith0 = as h approaches zero

Notations for Derivative

•f’(x) F prime•Dy Derivative of y in dx 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 f(x+h) – f(x) = lim x²+2xh=h²-x² = lim h(2x+h)

h0 h h0 h h0 h

When bottom H goes away, you can plug in 0 for h

Lim (2x+0) = 2x

h0

Thus, you can follow these three simple

steps!1. Replace x with x+h and copy problem2. Factor and get rid of h3. Replace h with 0

Let’s try one more!

Lim f(x+h) – f(x) = lim (x+h)^1/2-

x^1/2 h0 h h0 h

f(x)= √x Think of this problem as: f(x)= x^1/2

lim (x+h)^1/2-x^1/2 * (x+h)^1/2+x^1/2 h0 h *multiple by the conjugate (x+h)^1/2=x^1/2

lim x+h-x______ = lim ___1_______h0 h [(x+h)^1/2 + x^1/2] h0 (x+h)^1/2 + x^1/2= 1 ______ = 1____ =

1/2x^-1/2(x+0)^1/2 +x^1/2 2x^1/2

Old Rules to remember!

Lim sinh = 1h0 h

Lim 1-cosh = 0h0 h

*Sin(x+h) = sinxcosh + cosxsinh*Cos(x+h) = cosxcosh - sinxsinh

Before we move on…..

Trigonometry

f(x)=sinx f’(x)=cosx Simple steps, we learned earlier:

1. Replace x with x+h and copy problem2. Factor and get rid of h3. Replace h with 0

Now we are going to use these steps to prove f(x)=sinx f’(x)=cosx

F(x)=sinxf’(x)=cosx

Lim f(x+h) – f(x) = Lim sin(x+h)-sinx

h0 h h0 h

= Lim sinxcosh + cosxsinh- sinx = Lim sinx(cosh-1) + Lim cosxsinh

h0 h h0 h h0 h

= Lim sinx (0) + Lim cosx (1) = Lim cosx

h0 h0 h0

Now let’s prove that f(x)=cosx f’(x)=-

sinxLim f(x+h) – f(x) = Lim cos(x+h)-cosx

h0 h h0 h

= Lim cosxcosh - sinxsinh- cosx = Lim cosx(cosh-1) - Lim sinxsinh

h0 h h0 h h0 h

=Lim cosx (0) – Lim sinx(1) = -sinx

h0 h0

Now let’s prove that f(x)=tanx

f’(x)=sec²x*Think of f(x)=tanx as f(x)= sinx/cosxLim = sin(x+h) – sinx

H0 cos(x+h) cosx

h

*now you need to find a common denominator and then flip!

Lim sin(x+h) cosx – sinxcos(x+h) * _1_

H0 _____cos (x+h) cosx_______ h

__h__

1

Tangent continued…

Lim sin(x+h) cosx – sinxcos(x+h)

H0 h[cos(x+h) cosx]

Lim (sinxcosh + cosxsinh) * cosx- sinx (coscosh – sinxsinh)

H0 hcos(x+h) * cosx

Lim sinxcoshcosx + cos²xsinh – cosxcoshsinx + sin²xsinh*cancel out

H0 h cos (x+h) cosx

Tangent continued…

Lim sinh (sin²x + cos²x) = Lim 1 * 1_____

h0 h cos(x+h) cosx h0 cos(x+h)cosx

*side note: Never expand the bottom!!!

Lim 1______ = Lim 1___

h0 cos(x+0)cosx h0 cos²x

*which equals sec²x!!!

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 Rule2. Product Rule3. 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+100f’(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)²2. 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 Number5x-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)²Y’= 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) 1T2.cos(e^-x)A3.(e^-x)(-1)

Let’s Review real quick

y=(4x+1)²(1-x)³

Y=2-x 3x+1

Y=3x^(2/3)-4x^(½)-2

Try these!

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=251.2x+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/dx2y(dy/dx)=-2x3. 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

1. Multiplication addition2. Division subtraction3. Exponents multipliers

For example: y=2^x

Lny= xln2 *RULE: when you add the ln, the exponent comes down in front as so.

*now take the derivative of both sides!1 dy = x(0)+ln2(1)y dx Dy/dx=ln2(y)*plug in your original

equation for y Dy/dx=ln2(2^x)

Formula:

*y = a^xDy/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.

© Laura Berglind and Avery Gibson, 2011, AP Calculus AB, Autrey

THE END