Derivative Using Definition

[AvgStudent]

Find [imath]\boxed{\frac{d}{dx}e^{x^{2}}}[/imath] using definition of derivative.
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.

 

[Steven G]

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

 

[AvgStudent]

[math]\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}[/math]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]

 

[Deleted member 4993]

Find [imath]\boxed{\frac{d}{dx}e^{x^{2}}}[/imath] using definition of derivative.
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.

What is the Taylor's series expansion of e^x2

 

[topsquark]

What is the Taylor's series expansion of e^x2

You mean, what is the Taylor series for [imath]e^{x}[/imath]!

-Dan

 

[AvgStudent]

[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?

 

[topsquark]

[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?

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]

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!