using-the-1st-principle-find-the-derivative-of-y-ax-b-n-

Question Number 25845 by NECx last updated on 15/Dec/17

u s i n g t h e 1 s t p r i n c i p l e f i n d t h e

d e r i v a t i v e o f

y = {(a x + b)}^{n}

Answered by ajfour last updated on 15/Dec/17

\frac{d y}{d x} = \lim_{h \to 0} \frac{{(a x + a h + b)}^{n} - {(a x + b)}^{n}}{h}

= \lim_{h \to 0} \frac{a h}{h} \times \lim_{h \to 0} [{(a x + a h + b)}^{n - 1} +

{(a x + a h + b)}^{n - 2} (a x + b) +

{(a x + a h + b)}^{n - 3} {(a x + b)}^{2} + \dots

\dots . . + (a x + a h + b) {(a x + b)}^{n - 2} +

{(a x + b)}^{n - 1}]

\frac{d {(a x + b)}^{n}}{d x} = a n {(a x + b)}^{n - 1} .

Commented by NECx last updated on 16/Dec/17

i s t i l l d o n t g e t h o w y o u r e m o v e d

\lim_{x \to 0} \frac{a h}{h}

Commented by ajfour last updated on 16/Dec/17

x^{n} - y^{n} = (x - y) (x^{n - 1} + x^{n - 2} y + . .

\dots + {x y}^{n - 2} + y^{n - 1})

s o i n y o u r q u e s t i o n w e h a v e

x^{n} = {(a x + a h + b)}^{n}

a n d y^{n} = {(a x + b)}^{n}

s o (x - y) = a h .

Answered by rita1608 last updated on 15/Dec/17

\frac{l i m}{h \to 0} \frac{{(a (x + h) + b)}^{n} - {(a x + b)}^{n}}{h}

= \frac{l i m}{h \to 0} \frac{{(a x + a h + b)}^{n} - {(a x + b)}^{n}}{h}

= \frac{l i m}{h \to 0} \frac{{(a x + b)}^{n} - {(a x + b)}^{n}}{h} = 0

Commented by rita1608 last updated on 16/Dec/17

t h a n k s f o r t h e c o r r e c t i o n

Commented by prakash jain last updated on 15/Dec/17

In the last step you substituted

h = 0 in numerator and left

h in denominator . This is not correct .

with this you will get 0 for every

derivative .