Derivative Using Definition

AvgStudent

New member
Find [imath]\boxed{\frac{d}{dx}e^{x^{2}}}[/imath] using definition of derivative.
Hint: Using Taylor's series for [imath]e^\lambda[/imath]
$\frac{d}{dx}e^{x^{2}}= \lim_{h\to 0} \frac{e^{(x+h)^2}-e^{x^2}}{h}$Not sure how to start this one.

Jomo

Elite Member
Can you try to use the hint? Show us where you get from there.

AvgStudent

New member
$\lim_{h\to 0} \frac{e^{(x+h)^2}-e^{x^2}}{h}= \lim_{h\to 0} \frac{e^{(x^2+2xh+h^2)}-e^{x^2}}{h}= e^{x^2}\lim_{h\to 0} \frac{e^{h(2x+h)}-1}{h}$I tried to simplify the exponent and factored out [imath]e^{x^2}[/imath]. I'm not sure how to use the Taylor series to prove that the limit is [imath]2x[/imath]

Subhotosh Khan

Super Moderator
Staff member
Find [imath]\boxed{\frac{d}{dx}e^{x^{2}}}[/imath] using definition of derivative.
Hint: Using Taylor's series for [imath]e^\lambda[/imath]
$\frac{d}{dx}e^{x^{2}}= \lim_{h\to 0} \frac{e^{(x+h)^2}-e^{x^2}}{h}$Not sure how to start this one.
What is the Taylor's series expansion of ex2

topsquark

Senior Member
What is the Taylor's series expansion of ex2
You mean, what is the Taylor series for [imath]e^{x}[/imath]!

-Dan

AvgStudent

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

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topsquark

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

• AvgStudent

AvgStudent

New member
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
I see it. Thanks!