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Question Number 5715 by sanusihammed last updated on 24/May/16

$$Differentiate{x}^{x}fromthefirstprinciple.$$$$$$$$pleasehelp$$

Answered by Yozzii last updated on 25/May/16

$$Lety=x^x.Ifx>0wecanwrite$$$$lny={lnx}^x=xlnx.Keepinmindthat$$$$yisafunctionofx.Todifferentiate$$$$theequationwithrespecttox,weneedtofind$$$$\frac{d}{dx}\left(lny\right)and\frac{d}{dx}\left(xlnx\right).$$$$Letlny=u.Then,byfirstprinciples,$$$$\frac{du}{dy}=\underset{\delta y\to 0}{\lim }\frac{\delta u}{\delta y}=\underset{\delta y\to 0}{\lim }\frac{ln(y+\delta y)-lny}{\delta y}$$$$\frac{du}{dy}=\underset{\delta y\to 0}{\lim }\frac{ln(1+\frac{\delta y}{y})}{\delta y}=\underset{\delta y\to 0}{\lim }ln{(1+\frac{\delta y}{y})}^{1/\delta y}$$$$$$$$Letk=\delta y/y\Rightarrow \delta y=yk$$$$\frac{du}{dy}=\underset{k\to 0}{\lim }{(1+u)}^{1/\left(yk\right)}=\frac{1}{y}\left\{\underset{k\to 0}{\lim }ln{(1+k)}^{1/k}\right\}$$$$letk=1/n.$$$$\therefore \frac{du}{dy}=\frac{1}{y}\left\{\underset{n\to \mathrm{\infty }}{\lim }ln{(1+\frac{1}{n})}^n\right\}=\frac{1}{y}ln\left\{\underset{n\to \mathrm{\infty }}{\lim }{(1+n^{-1})}^n\right\}=\frac{1}{y}lne$$$$\Rightarrow \frac{du}{dy}=1/y.......\left(1\right)$$$$$$$$Now,\frac{du}{dx}=\frac{du}{dy}\times \frac{dy}{dx}accordingtothechainrule.$$$$PROOF:Fromfirstprinciples$$$$\frac{du}{dx}=\underset{\delta x\to 0}{\lim }\frac{\delta u}{\delta x}=\underset{\delta x\to 0}{\lim }\left(\frac{\delta u}{\delta y}\times \frac{\delta y}{\delta x}\right)=\left(\underset{\delta x\to 0}{\lim }\frac{\delta u}{\delta y}\right)\left(\underset{\delta x\to 0}{\lim }\frac{\delta y}{\delta x}\right)$$$$Ify=f\left(x\right)suchthat\delta y=f(x+\delta x)-f\left(x\right)$$$$\Rightarrow as\delta x\to 0,f(x+\delta x)-f\left(x\right)\to 0or\delta y\to 0.$$$$\therefore \underset{\delta x\to 0}{\lim }\frac{\delta u}{\delta y}=\underset{\delta y\to 0}{\lim }\frac{\delta u}{\delta y}=\frac{du}{dy}and\underset{\delta x\to 0}{\lim }\frac{\delta y}{\delta x}=\frac{dy}{dx}.$$$$\Rightarrow \frac{du}{dx}=\frac{du}{dy}\times \frac{dy}{dx}.◻$$$$$$$$\therefore \frac{du}{dx}=\frac{d\left(lny\right)}{dx}=\frac{d\left(lny\right)}{dy}\times \frac{dy}{dx}=\frac{1}{y}\bullet \frac{dy}{dx}.$$$$Now,$$$$\frac{d\left(xlnx\right)}{dx}=\underset{\delta x\to 0}{\lim }\frac{(x+\delta x)ln(x+\delta x)-xlnx}{\delta x}$$$$=\underset{\delta x\to 0}{\lim }\frac{xln(x+\delta x)-xlnx+\delta xln(x+\delta x)}{\delta x}$$$$=\underset{\delta x\to 0}{\lim }\{\frac{x}{\delta x}ln(1+\frac{\delta x}{x})+ln(x+\delta x)\}$$$$=x\left\{\underset{\delta x\to 0}{\lim }\frac{1}{\delta x}ln(1+\frac{\delta x}{x})\right\}+ln(x+0)$$$$Wejustfoundtheresultof\underset{\delta y\to 0}{\lim }\frac{1}{\delta y}ln(1+\frac{\delta y}{y})=\frac{1}{y}from\left(1\right).$$$$Hence,similarly\underset{\delta x\to 0}{\lim }\frac{1}{\delta x}ln(1+\frac{\delta x}{x})=\frac{1}{x}.$$$$\Rightarrow \frac{d\left(xlnx\right)}{dx}=x\times \frac{1}{x}+lnx=1+lnx(x>0)$$$$\therefore \frac{1}{y}\times \frac{dy}{dx}=1+lnx\Rightarrow \frac{dy}{dx}=y(1+lnx)=x^x(1+lnx)$$$$$$$$$$

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