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Question Number 52953 by hassentimol last updated on 15/Jan/19

$$$$$$\mathrm{Can}\mathrm{you}\mathrm{pls}\mathrm{help}\mathrm{me}\mathrm{with}\mathrm{that}\mathrm{but}\mathrm{only}$$$$\mathrm{by}\mathrm{using}\mathrm{the}\mathrm{derivative}\mathrm{of}f\left(1\right)\mathrm{and}\mathbf{not}$$$$\mathrm{using}\mathrm{the}\mathrm{defined}\mathrm{defivative}\mathrm{of}f\left(x\right)\mathrm{itself}?$$$$$$$$\mathrm{Thanks}$$$$$$$$f\left(x\right)=x^2-x$$$$$$$$f{}^{\prime }\left(1\right)=?$$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

$$f^{{}^{\prime }}\left(1\right)=\underset{h\to 0}{\lim }\frac{\{{(1+h)}^2-(1+h)\}-\{1^2-1\}}{h}$$$$=\underset{h\to 0}{\lim }\frac{1+2h+h^2-1-h}{h}$$$$=\underset{h\to 0}{\lim }\frac{h(1+h)}{h}$$$$=1+0=1ans$$

Commented by hassentimol last updated on 15/Jan/19

$$\mathrm{I}\mathrm{am}\mathrm{so}\mathrm{grateful}\mathrm{to}\mathrm{you}!\mathrm{I}\mathrm{was}\mathrm{asking}\mathrm{myself}$$$$\mathrm{why}\mathrm{my}\mathrm{answer}\mathrm{was}\mathrm{incorrect}\mathrm{and}\mathrm{now},$$$$\mathrm{thans}\mathrm{to}\mathrm{you},\mathrm{I}\mathrm{am}\mathrm{understamding}!$$

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

$$f^{,}\left(a\right)=\underset{h\to 0}{\lim }\frac{f(a+h)-f\left(a\right)}{h}thisisformula$$$$...thankyou...$$

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