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Question Number 16893 by chux last updated on 27/Jun/17

if y=u^n show that d^n y/dx^n =n!

$$\mathrm{if}\:\mathrm{y}=\mathrm{u}^{\mathrm{n}} \:\:\:\mathrm{show}\:\mathrm{that}\:\mathrm{d}^{\mathrm{n}} \mathrm{y}/\mathrm{dx}^{\mathrm{n}} =\mathrm{n}! \\ $$

Commented by mrW1 last updated on 28/Jun/17

if y=u^n then d^n y/dx^n =0 if y=x^n then d^n y/dx^n =n!

$$\mathrm{if}\:\mathrm{y}=\mathrm{u}^{\mathrm{n}} \:\:\:\mathrm{then}\:\mathrm{d}^{\mathrm{n}} \mathrm{y}/\mathrm{dx}^{\mathrm{n}} =\mathrm{0} \\ $$$$\mathrm{if}\:\mathrm{y}=\mathrm{x}^{\mathrm{n}} \:\:\:\mathrm{then}\:\mathrm{d}^{\mathrm{n}} \mathrm{y}/\mathrm{dx}^{\mathrm{n}} =\mathrm{n}! \\ $$

Commented by chux last updated on 28/Jun/17

please i dont really understand it

$$\mathrm{please}\:\mathrm{i}\:\mathrm{dont}\:\mathrm{really}\:\mathrm{understand}\:\mathrm{it} \\ $$

Commented by prakash jain last updated on 28/Jun/17

(d/dx)x^n =nx^(nāˆ’1) (d/dx)u^n =(d/du)u^n (du/dx)=nu^(nāˆ’1) (du/dx) If u is independent of x (du/dx)=0 (d/dx)u^n =0

$$\frac{{d}}{{dx}}{x}^{{n}} ={nx}^{{n}āˆ’\mathrm{1}} \\ $$$$\frac{{d}}{{dx}}{u}^{{n}} =\frac{{d}}{{du}}{u}^{{n}} \frac{{du}}{{dx}}={nu}^{{n}āˆ’\mathrm{1}} \frac{{du}}{{dx}} \\ $$$$\mathrm{If}\:{u}\:\mathrm{is}\:\mathrm{independent}\:\mathrm{of}\:{x}\:\frac{{du}}{{dx}}=\mathrm{0} \\ $$$$\frac{{d}}{{dx}}{u}^{{n}} =\mathrm{0} \\ $$

Commented by chux last updated on 28/Jun/17

thanks..... i get you now.

$$\mathrm{thanks}.....\:\mathrm{i}\:\mathrm{get}\:\mathrm{you}\:\mathrm{now}. \\ $$

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