Derive By First Principle – Example 1 Posted on July 11, 2019 by Ski Solving Equation of Index Number ExampleDerive the equation y=x2+3x by using first principle. Solution y=x2+3xdydx=limδx→0δyδx=limδx→0(y+δy)–yδx=limδx→0[(x+δx)2+3x+δx]–[x2+3x]δx=limδx→0x2+2xδx+δx2–x2+3x+δx–3xδx=limδx→02xδx+δx2δx+3x–3(x+δx)x(x+δx)δx=limδx→02x+δx+−3δxx(x+δx)×1δx=limδx→02x+δx+–3x(x+δx)=2x–3x2 Notes dydx=limδx→0δyδx
Selesaikan Persamaan Nombor Indeks – Contoh 1 Posted on July 10, 2019 by Ski Selesaikan Persamaan Nombor Indeks ContohSelesaikan persamaan 5x-1 + 5x+2 = 3150 Penyelesaian 5x–1+5x+2=31505x5+5x×52=31505x5+125×5x5=3150126×5x=5×31505x=5×31501265x=125=53x=3 Nota am×an=am+nam÷an=aman=am–nHence5x–1=5x÷5=5x55x+2=5x×52
Solving Equation of Index Number – Example 1 Posted on July 10, 2019 by Ski Solving Equation of Index Number ExampleSolve the equation 5x-1 + 5x+2 = 3150 Solution 5x–1+5x+2=31505x5+5x×52=31505x5+125×5x5=3150126×5x=5×31505x=5×31501265x=125=53x=3 Notes am×an=am+nam÷an=aman=am–nHence5x–1=5x÷5=5x55x+2=5x×52