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Question Number 45993 by peter frank last updated on 19/Oct/18

Commented by math khazana by abdo last updated on 20/Oct/18

$$\frac{d}{dx}\left(e^{x^2}\right)={lim}_{h\to 0}\frac{e^{{(x+h)}^2}-e^{x^2}}{h}$$$$={lim}_{h\to 0}\frac{e^{x^2+2xh+h^2}-e^{x^2}}{h}$$$$=e^{x^2}{lim}_{h\to 0}\frac{e^{2xh+h^2}-1}{h}but$$$$e^{2xh+h^2}=1+2xh+h^2+o\left(h^2\right)\Rightarrow$$$$\frac{e^{2xh+h^2}-1}{h}=2x+h+o\left(h\right)(h\to 0)\Rightarrow$$$${lim}_{h\to 0}\frac{e^{2xh+h^2}-1}{h}=2x\Rightarrow$$$$\frac{d}{dx}\left(e^{x^2}\right)=2x{e}^{x^2}.$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Oct/18

$$y=e^{x^2}y+△y=e^{{(x+△x)}^2}$$$$\frac{dy}{dx}=\underset{△x\to 0}{\lim }\frac{△y}{△x}$$$$=\underset{△x\to 0}{\lim }\frac{e^{{(x+△x)}^2}-e^{x^2}}{△x}$$$$=\underset{△x\to 0}{\lim }\frac{e^{x^2+2x△x+{(△x)}^2}-e^{x^2}}{△x}$$$$=\underset{△x\to 0}{\lim }\frac{e^{x^2}(e^{2x△x+{(△x)}^2}-1)}{△x}$$$$=\underset{△x\to 0}{\lim }{e}^{x^2}\times \frac{e^{2x△x+{(△x)}^2}-1}{2x△x+{(△x)}^2}\times \frac{2x+△x}{1}$$$$lett=2x△x+{(△x)}^{2}when△x\to 0t\to 0$$$$=e^{x^2}\times \frac{2x+0}{1}\times \underset{t\to 0}{\lim }\frac{e^t-1}{t}$$$$=e^{x^2}\times 2x\times 1=2{xe}^{x^2}isanswer$$

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