Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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

$$provethat$$$$\frac{d}{dx}{x}^n={nx}^{n-1}$$

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

$$\frac{d}{dx}{x}^n$$$$=\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{{(x+\mathrm{\Delta }x)}^n-x^n}{\mathrm{\Delta }x}$$$$=\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{x^n{(1+\frac{\mathrm{\Delta }x}{x})}^n-x^n}{\mathrm{\Delta }x}$$$$=\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{x^n[1+n\left(\frac{\mathrm{\Delta }x}{x}\right)+\frac{n(n-1)}{2!}{\left(\frac{\mathrm{\Delta }x}{x}\right)}^2+\frac{n(n-1)(n-2)}{3!}{\left(\frac{\mathrm{\Delta }x}{x}\right)}^3+...]-x^n}{\mathrm{\Delta }x}$$$$=\underset{\mathrm{\Delta }x\to 0}{\lim }{x}^n[n\left(\frac{1}{x}\right)+\frac{n(n-1)}{2!}\times \frac{\mathrm{\Delta }x}{x^2}+\frac{n(n-1)(n-2)}{3!}\times \frac{{\left(\mathrm{\Delta }x\right)}^2}{x^3}+...]$$$$=x^{n}n\left(\frac{1}{x}\right)$$$$={nx}^{n-1}$$

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

$$thankyousir$$

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

$$letf\left(x\right)=x^{n}wehave{f}^{{}^{\prime }}\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$$$$={lim}_{h\to 0}\frac{{(x+h)}^n-x^n}{h}={lim}_{h\to 0}\frac{(x+h-x)({(x+h)}^{n-1}+{(x+h)}^{n-2}x+....(x+h)x^{n-2}+x^{n-1}}{h}$$$$={lim}_{h\to 0}{(x+h)}^{n-1}+{(x+h)}^{n-2}x+....+(x+h)x^{n-2}+x^{n-1}$$$$=x^{n-1}+x^{n-1}+...+x^{n-1}\left(ntime\right)={nx}^{n-1}\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)={nx}^{n-1}$$

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

$$nicework.$$$$thankyousir$$

Commented by arcana last updated on 10/May/20

$$b$$$$\mathrm{the}\mathrm{correct}\mathrm{way}\mathrm{to}\mathrm{prove}\mathrm{it}\mathrm{is}\mathrm{for}$$$$\mathrm{induction}$$$$\ast \mathrm{prove}\mathrm{it}\mathrm{for}\mathrm{the}\mathrm{first}\mathrm{cases}.$$$$\ast \mathrm{assume}\mathrm{that}\mathrm{is}\mathrm{true},\mathrm{then}\mathrm{prove}\mathrm{the}$$$$\mathrm{case}\mathrm{n}+1.$$

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

$$youarewelcome.$$

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

$$\mathrm{correct}\mathrm{sir},\mathrm{but}\mathrm{the}\mathrm{method}\mathrm{used}\mathrm{above}\mathrm{can}$$$$\mathrm{also}\mathrm{be}\mathrm{used}\mathrm{since}\mathrm{the}\mathrm{question}\mathrm{didnt}\mathrm{specify}\mathrm{by}\mathrm{induction}.$$

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

$$\mathrm{by}\mathrm{mathematic}\mathrm{induction},$$$$\mathrm{prove}\mathrm{for}n=0:\frac{d}{dx}\left(x^0\right)=0\mathrm{and}n=0\mathrm{into}{nx}^{n-1}=0$$$$\mathrm{prove}\mathrm{for}n=1,\mathrm{LHS}=\frac{d}{dx}\left(x\right)=1$$$$\mathrm{RHS}=1{\left(x\right)}^{1-1}=1$$$$\Rightarrow \mathrm{the}\mathrm{derivative}\mathrm{formula}\mathrm{is}\mathrm{true}\mathrm{for}n=0,1$$$$\mathrm{assume}\mathrm{it}\mathrm{is}\mathrm{true}\mathrm{for}n=k$$$$\Rightarrow \frac{d}{dx}{x}^k={kx}^{k-1}$$$$\mathrm{prove}\mathrm{for}n=k+1$$$$\frac{d}{dx}{\left(x\right)}^{k+1}=\frac{d}{dx}\left(x^k\times x\right)=x^k+x\frac{d}{dx}\left(x^k\right)$$$$=x^k+x\left({kx}^{k-1}\right)$$$$=x^k+{kx}^k$$$$=(k+1)x^k$$$$\mathrm{hence}\mathrm{the}\mathrm{derivatve}\mathrm{formula}\mathrm{is}\mathrm{true}\mathrm{for}n=k+1$$$$\mathrm{hence}\mathrm{\forall }n\in \mathbb{Z},\frac{d}{dx}{x}^n={nx}^{n-1}.$$

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

$$nicework.$$$$thanx$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com