#### AvgStudent

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Hint: Using Taylor's series for [imath]e^\lambda[/imath]

[math]\frac{d}{dx}e^{x^{2}}= \lim_{h\to 0} \frac{e^{(x+h)^2}-e^{x^2}}{h}[/math]Not sure how to start this one.

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- Thread starter AvgStudent
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Hint: Using Taylor's series for [imath]e^\lambda[/imath]

[math]\frac{d}{dx}e^{x^{2}}= \lim_{h\to 0} \frac{e^{(x+h)^2}-e^{x^2}}{h}[/math]Not sure how to start this one.

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What is the Taylor's series expansion of e

Hint:Using Taylor's seriesfor [imath]e^\lambda[/imath]

[math]\frac{d}{dx}e^{x^{2}}= \lim_{h\to 0} \frac{e^{(x+h)^2}-e^{x^2}}{h}[/math]Not sure how to start this one.

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You mean, what is the Taylor series for [imath]e^{x}[/imath]!What is the Taylor's series expansion of e^{x2}

-Dan

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[math]e^{x^2}\lim_{h\to 0} \frac{e^{h(2x+h)}-1}{h}= e^{x^2}\lim_{h\to 0} \frac{\cancel{1}+\red{2xh}+h^2+\frac{1}{2!}(2xh+h^2)^2+\dots-\cancel{1}} {h}[/math]I spy with my little eyes the 2xh in red, so somehow get rid of everything else?

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After you divide by h then you have x + 2hx + ... so in the limit the second term (and higher in h) will go to 0.[math]e^{x^2}\lim_{h\to 0} \frac{e^{h(2x+h)}-1}{h}= e^{x^2}\lim_{h\to 0} \frac{\cancel{1}+\red{2xh}+h^2+\frac{1}{2!}(2xh+h^2)^2+\dots-\cancel{1}} {h}[/math]I spy with my little eyes the 2xh in red, so somehow get rid of everything else?

And [imath]f(x) \approx f(0) + f'(0)x + \dfrac{1}{2!}f''(0)x^2 + \text{ ...}[/imath] near x = 0, so for [imath]f(x) = e^{h(x + h)}[/imath] you get

[imath]e^{h(x + h)} \approx e^{h^2} + \left ( h e^{h^2} \right ) x + \dfrac{1}{2!} \left ( h^2 e^{h^2} \right ) x^2 + \text{ ...}[/imath]

-Dan

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I see it. Thanks!After you divide by h then you have x + 2hx + ... so in the limit the second term (and higher in h) will go to 0.

And [imath]f(x) \approx f(0) + f'(0)x + \dfrac{1}{2!}f''(0)x^2 + \text{ ...}[/imath] near x = 0, so for [imath]f(x) = e^{h(x + h)}[/imath] you get

[imath]e^{h(x + h)} \approx e^{h^2} + \left ( h e^{h^2} \right ) x + \dfrac{1}{2!} \left ( h^2 e^{h^2} \right ) x^2 + \text{ ...}[/imath]

-Dan