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Question-45993

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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
(d/dx)(e^x^2 )=lim_(h→0) ((e^((x+h)^2 ) −e^x^2 )/h) =lim_(h→0) ((e^(x^2 +2xh+h^2 ) −e^x^2 )/h) =e^x^2 lim_(h→0) ((e^(2xh+h^2 ) −1)/h) but e^(2xh+h^2 ) =1+2xh+h^2 +o(h^2 ) ⇒ ((e^(2xh+h^2 ) −1)/h) =2x +h +o(h) (h→0) ⇒ lim_(h→0) ((e^(2xh+h^2 ) −1)/h) =2x ⇒ (d/dx)(e^x^2 )=2x e^x^2 .
$$\frac{{d}}{{dx}}\left({e}^{{x}^{\mathrm{2}} } \right)={lim}_{{h}\rightarrow\mathrm{0}} \:\frac{{e}^{\left({x}+{h}\right)^{\mathrm{2}} } −{e}^{{x}^{\mathrm{2}} } }{{h}} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}} \:\frac{{e}^{{x}^{\mathrm{2}} +\mathrm{2}{xh}+{h}^{\mathrm{2}} } −{e}^{{x}^{\mathrm{2}} } }{{h}} \\ $$$$={e}^{{x}^{\mathrm{2}} } {lim}_{{h}\rightarrow\mathrm{0}} \:\frac{{e}^{\mathrm{2}{xh}+{h}^{\mathrm{2}} } −\mathrm{1}}{{h}}\:{but}\: \\ $$$${e}^{\mathrm{2}{xh}+{h}^{\mathrm{2}} } =\mathrm{1}+\mathrm{2}{xh}+{h}^{\mathrm{2}} \:+{o}\left({h}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\frac{{e}^{\mathrm{2}{xh}+{h}^{\mathrm{2}} } −\mathrm{1}}{{h}}\:=\mathrm{2}{x}\:+{h}\:+{o}\left({h}\right)\:\left({h}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${lim}_{{h}\rightarrow\mathrm{0}} \:\frac{{e}^{\mathrm{2}{xh}+{h}^{\mathrm{2}} } −\mathrm{1}}{{h}}\:=\mathrm{2}{x}\:\Rightarrow \\ $$$$\frac{{d}}{{dx}}\left({e}^{{x}^{\mathrm{2}} } \right)=\mathrm{2}{x}\:{e}^{{x}^{\mathrm{2}} } . \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Oct/18
y=e^(x^2 ) y+△y=e^((x+△x)^2 ) (dy/dx)=lim_(△x→0) ((△y)/(△x)) =lim_(△x→0) ((e^((x+△x)^2 ) −e^x^2 )/(△x)) =lim_(△x→0) ((e^(x^2 +2x△x+(△x)^2 ) −e^x^2 )/(△x)) =lim_(△x→0) ((e^x^2 (e^(2x△x+(△x)^2 ) −1))/(△x)) =lim_(△x→0) e^x^2 ×((e^(2x△x+(△x)^2 ) −1)/(2x△x+(△x)^2 ))×((2x+△x)/1) let t=2x△x+(△x)^2 when △x→0 t→0 =e^x^2 ×((2x+0)/1)×lim_(t→0) ((e^t −1)/t) =e^x^2 ×2x×1=2xe^x^2 is answer
$${y}={e}^{{x}^{\mathrm{2}} \:\:\:\:} \:{y}+\bigtriangleup{y}={e}^{\left({x}+\bigtriangleup{x}\right)^{\mathrm{2}} } \\ $$$$\frac{{dy}}{{dx}}=\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\bigtriangleup{y}}{\bigtriangleup{x}} \\ $$$$\:\:\:\:\:\:=\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{e}^{\left({x}+\bigtriangleup{x}\right)^{\mathrm{2}} } −{e}^{{x}^{\mathrm{2}} } }{\bigtriangleup{x}} \\ $$$$\:\:\:\:\:\:\:=\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{e}^{{x}^{\mathrm{2}} +\mathrm{2}{x}\bigtriangleup{x}+\left(\bigtriangleup{x}\right)^{\mathrm{2}} } −{e}^{{x}^{\mathrm{2}} } }{\bigtriangleup{x}} \\ $$$$\:\:\:\:\:\:\:\:=\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{e}^{{x}^{\mathrm{2}} } \left({e}^{\mathrm{2}{x}\bigtriangleup{x}+\left(\bigtriangleup{x}\right)^{\mathrm{2}} } −\mathrm{1}\right)}{\bigtriangleup{x}} \\ $$$$\:\:\:\:\:\:\:\:=\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:{e}^{{x}^{\mathrm{2}} } ×\frac{{e}^{\mathrm{2}{x}\bigtriangleup{x}+\left(\bigtriangleup{x}\right)^{\mathrm{2}} } −\mathrm{1}}{\mathrm{2}{x}\bigtriangleup{x}+\left(\bigtriangleup{x}\right)^{\mathrm{2}} }×\frac{\mathrm{2}{x}+\bigtriangleup{x}}{\mathrm{1}} \\ $$$${let}\:{t}=\mathrm{2}{x}\bigtriangleup{x}+\left(\bigtriangleup{x}\right)^{\mathrm{2}} \:\:\:{when}\:\bigtriangleup{x}\rightarrow\mathrm{0}\:\:\:\:{t}\rightarrow\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:={e}^{{x}^{\mathrm{2}} } ×\frac{\mathrm{2}{x}+\mathrm{0}}{\mathrm{1}}×\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{t}} −\mathrm{1}}{{t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:={e}^{{x}^{\mathrm{2}} } ×\mathrm{2}{x}×\mathrm{1}=\mathrm{2}{xe}^{{x}^{\mathrm{2}} } \:\:{is}\:{answer} \\ $$

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