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 + Δ_x) = (x + Δ_x)² + 4(x + Δ_x)
   = x² + 2·xΔ_x + Δ_x² + 4x + 4Δ_x
   
   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.
For x² (all multiply!) :
• take the power of x down;
• derivate x;
• and reduce the power of x.
  it gives you: 2 · (x)' · x^{2 − 1}  |  (x)' = 1
  ⇒ (x²)' = 2x

For 4x :

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


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’ u²
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	x²	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·x^2-1 + 3 = 4x + 3.



2. What is the derivative of 5z² + y³ − 7x⁴, from first principle?
   
   Answer:   y' = 5·2·z^2-1 + 3·y^3-1 − 7·4·x^4-1 = 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

   x²

   = 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 y   when y = (x² − 5x + 7)4