prove-that-d-dx-x-n-nx-n-1-

Question Number 93064 by M±th+et+s last updated on 10/May/20

p r o v e t h a t

\frac{d}{d x} x^{n} = {n x}^{n - 1}

Commented by mr W last updated on 10/May/20

\frac{d}{d x} x^{n}

= \lim_{Δ x \to 0} \frac{{(x + Δ x)}^{n} - x^{n}}{Δ x}

= \lim_{Δ x \to 0} \frac{x^{n} {(1 + \frac{Δ x}{x})}^{n} - x^{n}}{Δ x}

= \lim_{Δ x \to 0} \frac{x^{n} [1 + n (\frac{Δ x}{x}) + \frac{n (n - 1)}{2!} {(\frac{Δ x}{x})}^{2} + \frac{n (n - 1) (n - 2)}{3!} {(\frac{Δ x}{x})}^{3} + \dots] - x^{n}}{Δ x}

= \lim_{Δ x \to 0} x^{n} [n (\frac{1}{x}) + \frac{n (n - 1)}{2!} \times \frac{Δ x}{x^{2}} + \frac{n (n - 1) (n - 2)}{3!} \times \frac{{(Δ x)}^{2}}{x^{3}} + \dots]

= x^{n} n (\frac{1}{x})

= {n x}^{n - 1}

Commented by M±th+et+s last updated on 10/May/20

t h a n k y o u s i r

Commented by mathmax by abdo last updated on 10/May/20

l e t f (x) = x^{n} w e h a v e f^{^{'}} (x) = {l i m}_{h \to 0} \frac{f (x + h) - f (x)}{h}

= {l i m}_{h \to 0} \frac{{(x + h)}^{n} - x^{n}}{h} = {l i m}_{h \to 0} \frac{(x + h - x) ({(x + h)}^{n - 1} + {(x + h)}^{n - 2} x + \dots . (x + h) x^{n - 2} + x^{n - 1}}{h}

= {l i m}_{h \to 0} {(x + h)}^{n - 1} + {(x + h)}^{n - 2} x + \dots . + (x + h) x^{n - 2} + x^{n - 1}

= x^{n - 1} + x^{n - 1} + \dots + x^{n - 1} (n t i m e) = {n x}^{n - 1} \Rightarrow

f^{^{'}} (x) = {n x}^{n - 1}

Commented by M±th+et+s last updated on 10/May/20

n i c e w o r k .

t h a n k y o u s i r

Commented by arcana last updated on 10/May/20

b

the correct way to prove it is for

induction

* prove it for the first cases .

* assume that is true, then prove the

case n + 1 .

Commented by mathmax by abdo last updated on 11/May/20

y o u a r e w e l c o m e .

Commented by Rio Michael last updated on 11/May/20

correct sir, but the method used above can

also be used since the question didnt specify by induction .

Answered by Rio Michael last updated on 11/May/20

by mathematic induction,

prove for n = 0 : \frac{d}{d x} (x^{0}) = 0 and n = 0 into {n x}^{n - 1} = 0

prove for n = 1, LHS = \frac{d}{d x} (x) = 1

RHS = 1 {(x)}^{1 - 1} = 1

\Rightarrow the derivative formula is true for n = 0, 1

assume it is true for n = k

\Rightarrow \frac{d}{d x} x^{k} = {k x}^{k - 1}

prove for n = k + 1

\frac{d}{d x} {(x)}^{k + 1} = \frac{d}{d x} (x^{k} \times x) = x^{k} + x \frac{d}{d x} (x^{k})

= x^{k} + x ({k x}^{k - 1})

= x^{k} + {k x}^{k}

= (k + 1) x^{k}

hence the derivatve formula is true for n = k + 1

hence \forall n \in Z, \frac{d}{d x} x^{n} = {n x}^{n - 1} .

Commented by M±th+et+s last updated on 11/May/20

n i c e w o r k .

t h a n x