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

Guest Oct 1, 2014

Best Answer 

 #2
avatar+85726 
+5

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)xn-1 Δx + C(n,2) xn-2 Δx2 + ..... +C(n, n-1) xΔxn-1 + C(n, n)Δxn  - xn ] / (Δx) 

Δx → 0

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

Δx → 0   

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

lim         (Δx) [C(n, 1) xn-1  +C(n, 2) xn-2 Δx + ..... +C(n, n-1) xΔxn-2 +C (n, n) Δxn-1 ] / (Δx)

Δx → 0

lim          [C(n, 1) xn-1  + C(n, 2) xn-2 Δx + ..... +C(n, n-1) xΔxn-2 + C(n, n) Δxn-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......

nxn-1

 

CPhill  Oct 1, 2014
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2+0 Answers

 #1
avatar+17721 
+5

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 )

geno3141  Oct 1, 2014
 #2
avatar+85726 
+5
Best Answer

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)xn-1 Δx + C(n,2) xn-2 Δx2 + ..... +C(n, n-1) xΔxn-1 + C(n, n)Δxn  - xn ] / (Δx) 

Δx → 0

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

Δx → 0   

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

lim         (Δx) [C(n, 1) xn-1  +C(n, 2) xn-2 Δx + ..... +C(n, n-1) xΔxn-2 +C (n, n) Δxn-1 ] / (Δx)

Δx → 0

lim          [C(n, 1) xn-1  + C(n, 2) xn-2 Δx + ..... +C(n, n-1) xΔxn-2 + C(n, n) Δxn-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......

nxn-1

 

CPhill  Oct 1, 2014

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