Montag, 26. Januar 2015

The Derivative from First Principles


Derivative Rules , Common Functions , Tricks , Examples , Exercises ,

Definition:

From first principles is also known as differentiating a function by starting from scratch (Zero) and use algebra to find a general expression for the slope of a curve, at any x value. First principles use the "delta method" Δx (change in x) and Δy (change in y).


Remember:
The Slope is defined as:  m  = 
y2 − y1
x2 − x1
 = 
Δy
Δx
 = 
f(x + Δx) − f(x)
Δx
 = 
change in y
change in y
The exact Slope of a curve is found at its tangent by using the derivative or differentiation.

Putting together, the Slope of a tangent can be written as follow:

dy
dx
 =  lim
Δx → 0
 
Δy
Δx
 =  lim
Δx → 0
 
f(x + Δx) − f(x)
Δx
With y = f(x), the derivative y'  =  f(x)'  =  
dy
dx

Example:
  1. Find y' from first principles if y = x2 + 4x
  2. Find the slope of the tangent where x = 1 and also where x = − 6.
  3. Sketch the curve and both tangents.

Answer:
  1. Note: y' means "the first derivative", it can also be written
    dy
    dx
    Assuming f(x) = x2 + 4x

    f(x + Δx) = (x + Δx)² + 4(x + Δx)
    = x² + 2·xΔx + Δx² + 4x + 4Δx

    Therefore

    dy
    dx
     =  lim
    Δx → 0
     
    (x² + 2·xΔx + Δx² + 4x + 4Δx) − (x² + 4x)
    Δx

     =  lim
    Δx → 0
      (2x + Δx + 4)
     =  2x + 4.

    From first principles y' = 2x + 4.



  2. When x = 1, slope m = 2(1) + 4 = 6.

    When x = −6, slope m = 2(-6) + 4 =  − 8.


  3. Sketch:


Some tricks!!!

Find y' from first principles if y = x² + 4x
    To find the derivative of the function y = f(x): do it separately as is an addition of two functions.
  1. For x² (all multiply!) :
    • take the power of x down;
    • derivate x;
    • and reduce the power of x.
    •   it gives you: 2 · (x)' · x2 − 1  |  (x)' = 1
        ⇒ (x²)' = 2x

    For 4x :
    • 4 ist a constant and can be derivated;
    • the derivative of x, (x)' = 1
    •   ⇒ (4x)' = 4

  2. Now add both: y' = 2x + 4, and that's the derivative of the function y = f(x).


Rules Function Derivative
Multiplication by constant c·u c·u’
Sum Rule u + v u’ + v’
Difference Rule u − v u’ − v’
Power Rule xn n·xn − 1
Product Rule u · v u · v’ + u’· v
Quotient Rule
u
v
u’·v − v’·u
v2
Reciprocal Rule
1
u
−u’
Chain Rule
(in a different form)
u(v(x)) u’(v(x))v’(x)
Chain Rule
(as "Composition of Functions")
u○v (u’○v)·v'


Common Functions Function Derivative
Constant c 0
Straight Function
Linear Function
x 1
Square Function 2·x
Square Root x ¹/2 · x-1/2
Exponential ex ex
Logarithms ax

ln(x)

loga(x)
ax(ln a)
1
x
1
(x ln(a))
Trigonometry
(x is in radians)
sin(x)

cos(x)

tan(x)

sin-1(x)

cos-1(x)

tan-1(x)
cos(x)

−sin(x)

1/cos²x

-cos(x)/sin²x

sin(x)/cos²x

1/1 + x²



Examples


  1. ​ ​
    Find y = 
    dy
    dx
    from first principles if y = 2x² + 3x.
    Answer:   y' = 2 · 2·x2-1 + 3 = 4x + 3.


  2. What is the derivative of 5z² + y3 − 7x4, from first principle?

    Answer:   y' = 5·2·z2-1 + 3·y3-1 − 7·4·x4-1 = 10z + 3y² − 28x3


  3. ​ ​
    What is
    d
    dx
    ((x − 1)(x + 3))?

    Answer: 2 methods are possible!
    • Method 1: We can use the product rule
    • assuming that:
                          u = x − 1     ==>     u' = 1
                          v = x + 3     ==>     v' = 1

      Cross multiply and add them as followed:

      u·v = u'·v + u·v' = 1·(x + 3) + (x − 1)·1

      = x + 3 + x − 1

      = 2x + 2 = 2(x + 1)

    • Method 2: First expand the terms by multiplying out (remove the parentheses) and then use the Sum or Difference Rule to differentiate it separately

    • (x − 1)(x + 3) = x² + 3x − x − 3
      = x² + 2x − 3
      *
      d
      dx
      = 2x
      *
      d
      dx
      (2x) = 2
      *
      d
      dx
      (− 3) = 0
      Hence, (x − 1)(x + 3) = 2x + 2 = 2(x + 1)



Exercises


Differentiate from first principles:
i. y = x² − 4x
ii.
f(x) =
1
x
iii.
Find 
d
x
 when y = (x² − 5x + 7)4

Sonntag, 11. Januar 2015

The Binomial Theorem: Formulas

The formal expression of the Binomial Theorem is as follows:

(a + b)n =

Σ

(
n
k
) an − k bk

with  (
n
k
)  =  nck  = 
n!
(n−k)!·k!
  (often called "n, choose k")

The factorial notation "n!" means " the product of all the whole numbers between 1 and n".