Derive By First Principle – Example 1

Solving Equation of Index Number

Example

Derive the equation y=x2+3x by using first principle.

y=x2+3xdydx=limδx0δyδx=limδx0(y+δy)yδx=limδx0[(x+δx)2+3x+δx][x2+3x]δx=limδx0x2+2xδx+δx2x2+3x+δx3xδx=limδx02xδx+δx2δx+3x3(x+δx)x(x+δx)δx=limδx02x+δx+3δxx(x+δx)×1δx=limδx02x+δx+3x(x+δx)=2x3x2

dydx=limδx0δyδx

Selesaikan Persamaan Nombor Indeks – Contoh 1

Selesaikan Persamaan Nombor Indeks

Contoh

Selesaikan persamaan 5x-1 + 5x+2 = 3150

5x1+5x+2=31505x5+5x×52=31505x5+125×5x5=3150126×5x=5×31505x=5×31501265x=125=53x=3

am×an=am+nam÷an=aman=amnHence5x1=5x÷5=5x55x+2=5x×52

Solving Equation of Index Number – Example 1

Solving Equation of Index Number

Example

Solve the equation 5x-1 + 5x+2 = 3150

5x1+5x+2=31505x5+5x×52=31505x5+125×5x5=3150126×5x=5×31505x=5×31501265x=125=53x=3

am×an=am+nam÷an=aman=amnHence5x1=5x÷5=5x55x+2=5x×52