If-y-x-1-3-Find-dy-dx-from-the-first-principle-

Question Number 13068 by tawa tawa last updated on 13/May/17

If y = \sqrt[3]{x} Find \frac{dy}{dx} from the first principle

Answered by ajfour last updated on 13/May/17

\frac{d y}{d x} = \lim_{h \to 0} \frac{{(x + h)}^{1 / 3} - x^{1 / 3}}{h}

= \lim_{h \to 0} \frac{x^{1 / 3} {{(1 + \frac{h}{x})}^{1 / 3} - 1}}{h}

a s h ≪ x w e h a v e {(1 + \frac{h}{x})}^{n} = 1 + \frac{n h}{x}

s o \frac{d y}{d x} = \lim_{h \to 0} \frac{x^{1 / 3} {1 + \frac{h}{3 x} - 1}}{h}

\Rightarrow \frac{d y}{d x} = (x^{1 / 3}) \lim_{h \to 0} \frac{(h / 3 x)}{h}

= x^{1 / 3} (\frac{1}{3 x})

= \frac{1}{3 x^{2 / 3}} .

Commented by tawa tawa last updated on 13/May/17

God bless you sir

Commented by Nayon last updated on 23/May/17

w h y {(1 + \frac{h}{x})}^{n} = 1 + \frac{n h}{x} ?

Answered by mrW1 last updated on 13/May/17

G e n e r a l l y f o r y = f (x) = x^{a} (a \neq 0) w e h a v e

f (x + Δ x) = f (x + h) = {(x + h)}^{a}

\frac{Δ y}{Δ x} = \frac{f (x + Δ x) - f (x)}{Δ x} = \frac{{(x + h)}^{a} - x^{a}}{h} = x^{a} \frac{{(1 + \frac{h}{x})}^{a} - 1}{h}

\frac{d y}{d x} = \lim_{h \to 0} \frac{Δ y}{Δ x} = x^{a} \times \lim_{h \to 0} \frac{{(1 + \frac{h}{x})}^{a} - 1}{h}

{(1 + \frac{h}{x})}^{a} = 1 + a (\frac{h}{x}) + \frac{a (a - 1)}{2!} {(\frac{h}{x})}^{2} + \frac{a (a - 1) (a - 2)}{3!} {(\frac{h}{x})}^{3} + \cdot \cdot \cdot

\frac{{(1 + \frac{h}{x})}^{a} - 1}{h} = \frac{a}{x} + \frac{a (a - 1) h}{2! x^{2}} + \frac{a (a - 1) (a - 2) h^{2}}{3! x^{3}} + \cdot \cdot \cdot

\lim_{h \to 0} \frac{{(1 + \frac{h}{x})}^{a} - 1}{h} = \frac{a}{x} + 0 + 0 + 0 + \cdot \cdot \cdot = \frac{a}{x}

\Rightarrow \frac{d y}{d x} = x^{a} \times \frac{a}{x} = {a x}^{a - 1}

f o r a = \frac{1}{3} w e g e t

\frac{d y}{d x} = \frac{1}{3} x^{- \frac{2}{3}}

Commented by ajfour last updated on 13/May/17

t o o g o o d!

Commented by tawa tawa last updated on 13/May/17

God bless you sir .