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Question Number 97972 by Aniruddha Ghosh last updated on 10/Jun/20

$$\mathrm{Prove}\mathrm{that},$$$$\frac{\mathit{d}}{\mathit{d}\mathit{x}}\left({\mathit{e}}^{\mathit{x}}\right)={\mathit{e}}^{\mathit{x}}$$

Answered by abdomathmax last updated on 10/Jun/20

$$\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{\mathrm{x}}\right)={\lim }_{\mathrm{h}\to 0}\frac{{\mathrm{e}}^{\mathrm{x}+\mathrm{h}}-{\mathrm{e}}^{\mathrm{x}}}{\mathrm{h}}$$$$={\lim }_{\mathrm{h}\to 0}{\mathrm{e}}^{\mathrm{x}}\times \frac{{\mathrm{e}}^{\mathrm{h}}-1}{\mathrm{h}}$$$$\mathrm{we}\mathrm{have}{\mathrm{e}}^{\mathrm{h}}=1+\mathrm{h}+\mathrm{o}\left({\mathrm{h}}^2\right)\Rightarrow {\mathrm{e}}^{\mathrm{h}}-1=\mathrm{h}+\mathrm{o}\left({\mathrm{h}}^2\right)\Rightarrow$$$$\frac{{\mathrm{e}}^{\mathrm{h}}-1}{\mathrm{h}}=1+\mathrm{o}\left(\mathrm{h}\right)\Rightarrow {\lim }_{\mathrm{h}\to 0}\frac{{\mathrm{e}}^{\mathrm{h}}-1}{\mathrm{h}}=1\Rightarrow$$$$\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{\mathrm{x}}\right)={\mathrm{e}}^{\mathrm{x}}$$

Answered by MJS last updated on 10/Jun/20

$$\frac{d}{dx}\left[f\left(x\right)\right]=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x-h)}{2h}$$$$\frac{d}{dx}\left[{\mathrm{e}}^x\right]=\underset{h\to 0}{\lim }\frac{{\mathrm{e}}^{x+h}-{\mathrm{e}}^{x-h}}{2h}=\underset{h\to 0}{\lim }\frac{{\mathrm{e}}^h-{\mathrm{e}}^{-h}}{2h}{\mathrm{e}}^x=$$$$={\mathrm{e}}^x\times \underset{h\to 0}{\lim }\frac{\sinh h}{h}=$$$$\left[{\mathrm{l}}^{\prime }\mathrm{Hopital}\right]$$$$={\mathrm{e}}^x\times \underset{h\to 0}{\lim }\frac{\frac{d}{dh}\left[\sinh h\right]}{\frac{d}{dh}\left[h\right]}={\mathrm{e}}^x\times \underset{h\to 0}{\lim }\cosh h={\mathrm{e}}^x$$

Answered by smridha last updated on 10/Jun/20

$$\left(\mathit{i}\right)\frac{\mathit{d}}{\mathit{d}\mathit{x}}\left({\mathit{e}}^{\mathit{x}}\right)=\underset{\mathit{h}\to 0}{\lim }\frac{{\mathit{e}}^{\mathit{x}+\mathit{h}}-{\mathit{e}}^{\mathit{x}}}{\mathit{h}}\left[\mathit{j}\mathit{u}\mathit{s}\mathit{t}\mathit{u}\mathit{s}\mathit{i}\mathit{n}\mathit{g}\mathit{t}\mathit{h}\mathit{e}\mathit{d}\mathit{e}\mathit{f}\mathit{i}\mathit{n}\mathit{e}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\right]$$$$[\mathit{t}\mathit{h}\mathit{a}\mathit{t}\frac{\mathit{d}\mathit{f}}{\mathit{d}\mathit{x}}=\underset{\mathit{h}\to 0}{\lim }\frac{\mathit{f}(\mathit{x}+\mathit{h})-\mathit{f}\left(\mathit{x}\right)}{\mathit{h}}]$$$$\mathit{s}\mathit{o}\mathit{w}\mathit{e}\mathit{g}\mathit{e}\mathit{t}:$$$$={\mathit{e}}^{\mathit{x}}\left[\underset{\mathit{h}\to 0}{\lim }\frac{{\mathit{e}}^{\mathit{h}}-1}{\mathit{h}}\right]={\mathit{e}}^{\mathit{x}}.1={\mathit{e}}^{\mathit{x}}.$$$$\left(\mathit{i}\mathit{i}\right)\mathit{n}\mathit{o}\mathit{w}\mathit{d}\mathit{o}\mathit{i}\mathit{t}\mathit{b}\mathit{y}\mathit{s}\mathit{e}\mathit{r}\mathit{i}\mathit{e}\mathit{s}\mathit{e}\mathit{x}\mathit{p}\mathit{a}\mathit{n}\mathit{s}\mathit{s}\mathit{i}\mathit{o}\mathit{n}$$$${\mathit{e}}^{\mathit{x}}=\underset{\mathit{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathit{x}}^{\mathit{n}}}{\mathit{n}!}=1+\frac{\mathit{x}}{1!}+\frac{{\mathit{x}}^2}{2!}+\frac{{\mathit{x}}^3}{3!}+.....+\frac{{\mathit{x}}^{\mathit{n}}}{\mathit{n}!}+\frac{{\mathit{x}}^{\mathit{n}+1}}{(\mathit{n}+1)!}+....\mathrm{\infty }$$$$\mathit{n}\mathit{o}\mathit{w}\mathit{d}\mathit{i}\mathit{f}\mathit{f}:\mathit{w}\mathit{r}\mathit{t}\mathit{x}\mathit{w}\mathit{e}\mathit{g}\mathit{e}\mathit{t}$$$$\frac{\mathit{d}}{\mathit{d}\mathit{x}}\left({\mathit{e}}^{\mathit{x}}\right)=0+[1+\frac{\mathit{x}}{1!}+\frac{{\mathit{x}}^2}{2!}+\frac{{\mathit{x}}^3}{3!}+...+\frac{x^{\mathit{n}-1}}{(\mathit{n}-1)!}+\frac{{\mathit{x}}^{\mathit{n}}}{\mathit{n}!}+...\mathrm{\infty }]$$$$=0+{\mathit{e}}^{\mathit{x}}={\mathit{e}}^{\mathit{x}}\left[\mathit{i}\mathit{t}\mathit{s}\mathit{r}\mathit{e}\mathit{m}\mathit{a}\mathit{i}\mathit{n}\mathit{t}\mathit{h}\mathit{e}\mathit{a}\mathit{s}\mathit{i}\mathit{t}\mathit{i}\mathit{s}\right]$$

Commented by MathGod last updated on 10/Jun/20

$$\mathbb{I}\mathbb{USE}\mathbb{THE}\mathbb{REDDEST}\mathbb{INK}!$$

Commented by arcana last updated on 11/Jun/20

$$\mathrm{using}{e}^h-1-h⩾0,\mathrm{\forall }h\in \mathbb{R}$$$$\mathrm{then}1⩽\frac{e^h-1}{h}⩽1$$$$\underset{h\to 0}{\lim }\frac{e^h-1}{h}=1$$$$$$

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