• The solution to Newton's GEXACT VALUE & FORMULA
  • The pressure at the centre of a massTHE MATHEMATICAL LAW (calculator)
  • The pressure at the centre of a massEARTH'S INTERNAL STRUCTURE (calculator)
  • Proof of the non-exitence of Dark MatterDOES NOT EXIST
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The solution to Newton's G1 The theory controlling planetary spin2 Pressure at the centre of the Earth3 Proof of the non-exitence of Dark Matter4 The atom as Newton describes it5
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Differential Function, Product & Quotient Rules

The following table contains some rules and worked examples for differentiation.

y = a

dy/dx = 0
Because 'a' is a constant, the slope of the equation (dy/dx) must be zero

y = a.xn

dy/dx = n.a.xn-1

y = a.x + b.xm + c.xn

dy/dx = ℓ.a.xℓ-1 + m.b.xm-1 + n.c.xn-1

y = u + v

dy/dx = du/dx + dv/dx

Function of a Function
(algebraic)

y = (axn - b)m
u = axn – b > du/dx = a.n.xn-1
y = um > dy/du = m.um-1
dy/dx = dy/du . du/dx

Function of a Function
(trigonometric)

y = Sin²(a + bx²)
y = u² > dy/du = 2u
u = Sin(v) > du/dv = Cos(v)
v = a + bx² > dv/dx = 2bx
dy/dx = dy/du . du/dv . dv/dx
dy/dx = 2u . Cos(v) . 2.bx = 2.Sin(v) . Cos(a + bx²) . 2bx
dy/dx = 2.Sin(a + bx²) . Cos(a + bx²) . 2bx
dy/dx = 2bx . Sin(2(a + bx²))

Product Rule

y = u.v
dy/dx = u.dv/dx + v.du/dx

Product Rule
(one product)

y = (3x² + 2x).( 6x – 2x³)
u = 3x² + 2x
v = 6x – 2x³
y = u.v
dy/dx = u.dv/dx + v.du/dx dy/dx = d(3x² + 2x).(6x – 2x³)/dx + d(6x – 2x³).(3x² + 2x)/dx
dy/dx = (6x + 2).(6x – 2x³) + (6 – 6x).(3x² + 2x)
dy/dx = (36x² – 12x⁴ + 12x – 4x³) + (18x² + 12x – 18x³ – 12 x2)
dy/dx = -12x⁴ – 4x³ – 18x³ + 36x² + 18x² – 12x2 + 12x + 12x
dy/dx = -12x⁴ – 22x³ + 42x² +24x

Product Rule
(two products)

y = (1 + x²).(2 – x²).(3 + x²)
u = 1 + x²
v = 2 – x²
w = 3 + x²
y = u.v.w
dy/dx = u.v.dw/dx + u.w.dv/dx + v.w.du/dx
dy = (1 + x²).(2 – x²).d(3 + x²)/dx + (1 + x²).(3 + x²).d(2 – x²)/dx + (2 – x²).(3 + x²).d(1 + x²)/dx
dy/dx = (1 + x²).(2 – x²).(2x) + (1 + x²).(3 + x²).(2x) + (2 – x²).(3 + x²).(2x)
dy/dx = 2x.(2 – x⁴ + 2x² – x²) + 2x.(3 + x⁴ + 3x² + x²) + 2x.(6 – x⁴ + 2x² – 3x²)
dy/dx = (4x – 2x⁵ + 4x³ – 2x³) + (6x + 2x⁵ + 6x³ + 2x³) + (12x – 2x⁵ + 4x³ – 6x³)
dy/dx = 4x + 6x + 12x + 4x³ – 2x³ – 6x³ + 2x³ + 4x³ + 6x³ - 2x⁵ + 2x⁵ – 2x⁵
dy/dx = 22x + 8x³ - 2x⁵

Quotient Rule

y = u/v
dy/dx = (v.du/dx - u.dv/dx) / v²
y = 3x / (x - 1)
u = 3x
v = x - 1
dy/dx = [3(x-1) - 3x(1)] / (x² - 2x + 1)
dy/dx = (3x - 3 - 3x) / (x² - 2x + 1)
dy/dx = -3 / (x² - 2x + 1)

Partial Differentiation

2.x³ – 3.xy + 5.x – 4.y² + 3 = 0
6.x²,    3.x.y′ + 3.y,    5,    8.y.y′,    0
6.x² – [3.x.y′ + 3.y] + 5 – 8.y.y′ + 0 = 0
6.x² – 3.x.y′ – 3.y + 5 – 8.y.y′ + 0 = 0
3.x.y′ + 8.y.y′ = 6.x² – 3.y + 5
y′(3.x + 8.y) = 6.x² – 3.y + 5
y′ = (6.x² – 3.y + 5) / (3.x + 8.y)

Further Reading

You will find further reading on this subject in reference publications(19)

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