How do you derive the power rule that is used in calculus for differentiation?

We can also find this by using the difference quotient and the binomial theorem......suppose we want to find the derivative of xⁿ   ...... so we have.....

lim          [ (x + Δx)ⁿ - xⁿ ] / (Δx)   Δx → 0

And by the binomial theorem, we can write

lim          [ xⁿ + C(n, 1)x^n-1 Δx + C(n,2) x^n-2 Δx² + ..... +C(n, n-1) xΔx^n-1 + C(n, n)Δxⁿ  - xⁿ ] / (Δx) 

Δx → 0

lim          [C(n,1) x^n-1 Δx + C(n,2)x^n-2 Δx² + ..... +C(n, n-1) xΔx^n-1 +C(n, n) Δxⁿ ] / (Δx)      

Δx → 0   

.... factor out Δx in the numerator...    

lim         (Δx) [C(n, 1) x^n-1  +C(n, 2) x^n-2 Δx + ..... +C(n, n-1) xΔx^n-2 +C (n, n) Δx^n-1 ] / (Δx)

Δx → 0

lim          [C(n, 1) x^n-1  + C(n, 2) x^n-2 Δx + ..... +C(n, n-1) xΔx^n-2 + C(n, n) Δx^n-1 ]

Δx → 0

And taking the limit as Δx → 0 we have

C(n,1) x ^n-1   ..........  but   C(n, 1) is just n......so we have......

nx^n-1

Start by writing  x^n as e^(n lnx ).

d( x^n )/dx  =  d( e^(n lnx ) )/dx

                   =  e^( n ln x ) · d( n ln x )/dx                        chain rule

                    = x^n · ( n/x )

                    = n · x^( n - 1 )