A lil toughy: d/dx(sqrt(1+e^x)) from first principles

[etotheipi]

Any help will be greatly appreciated.

Can anyone differentiate this (from first principles):

sqrt( 1 + e^x)

Thanks

 

[stapel]

Can anyone differentiate this (from first principles)....

Probably. The real question is: Can you? :wink:

Which formula are you supposed to be using? (There is more than one, and there are different degrees of "first principles".) How far have you gotten?

Please be complete. Thank you!

Eliz.

 

[etotheipi]

ok sorry sorry, i like your style

So im using the standard first principle formula:

My attempted method has been too expand the top two brackets with the binomial theorem. You can imagine the result.

Apparently I should use the identity a - b = (a^2 - b^2) / (a + b)

So I made the two series produced through expansion a and b, but still this just makes it worse.

Please help.

 

[stapel]

Apparently I should use the identity a - b = (a^2 - b^2) / (a + b)

That's a good idea. Let's work in pieces:

. . . . .\(\displaystyle f(x)\, =\, \sqrt{1\, +\, e^x}\)

. . . . .\(\displaystyle f(x\, +\, h)\, =\, \sqrt{1\, +\, e^{x+h}}\)

. . . . .\(\displaystyle f(x\, +\, h)\, -\, f(x)\, =\, \sqrt{1\, +\, e^{x+h}}\, -\, \sqrt{1\, +\, e^x}\)

Now apply the identity that was suggested, by multiplying, top and bottom, by the conjugate of the above:

. . . . .\(\displaystyle \left(\frac{\sqrt{1\, +\, e^{x+h}}\, -\, \sqrt{1\, +\, e^x}}{1}\right)\,\left(\frac{\sqrt{1\, +\, e^{x+h}}\, +\, \sqrt{1\, +\, e^x}}{\sqrt{1\, +\, e^{x+h}}\, +\, \sqrt{1\, +\, e^x}}\right)\)

. . . . .\(\displaystyle \frac{(1\, +\, e^{x+h})\, -\, (1\, +\, e^x)}{\sqrt{1\, +\, e^{x+h}}\, +\, \sqrt{1\, +\, e^x}}\)

. . . . .\(\displaystyle \frac{e^{x+h}\, -\, e^x}{\sqrt{1\, +\, e^{x+h}}\, +\, \sqrt{1\, +\, e^x}}\)

. . . . .\(\displaystyle \frac{e^h e^x - e^x}{\sqrt{1\, +\, e^h e^x}\, +\, \sqrt{1\, +\, e^x}}\)

. . . . .\(\displaystyle \frac{e^x (e^h - 1)}{\sqrt{1\, +\, e^h e^x}\, +\, \sqrt{1\, +\, e^x}}\)

Now include the "divided by h" part:

. . . . .\(\displaystyle \frac{e^x (e^h - 1)}{h(\sqrt{1\, +\, e^h e^x}\, +\, \sqrt{1\, +\, e^x})}\)

The actual derivative is:

. . . . .\(\displaystyle \frac{e^x}{2\sqrt{1\, +\, e^x}}\)

...so clearly we're close. (The e[sup:16v67bek]h[/sup:16v67bek] in the first radical in the denominator will go to 1 as h goes to zero, leaving you with the desired sum.) But I'm drawing a blank on how to "cancel" the e[sup:16v67bek]h[/sup:16v67bek] - 1 and the h to finish this off....

Eliz.

 

[galactus]

Let's cheat and multiply it by \(\displaystyle \frac{h}{e^{h}-1}\) and get rid of the \(\displaystyle \frac{e^{h}-1}{h}\)

 

[Dr. Flim-Flam]

lim h->0 [e^x(e^h-1)]/ h* { [(1+e^(x+h)]^(1/2) +[1+e^x]^(1/2) } = lim [(e^h-1)/h]* lim e^x/{ [(1+e^(x+h)]^(1/2)+[1+e^x]}
h->0 h->0

lim [(e^h-1)/h = 1, hence 1* lim e^x/ {[(1+e^(x+h))^(1/2)] + [(1+e^x)^(1/2)]} = e^x/[2(1+e^x)^(1/2)]
h->0 h->0

 

[galactus]

Here's a way to find \(\displaystyle \lim_{h\to{0}}\frac{e^{h}-1}{h}\) without L'Hopital.

Let \(\displaystyle \frac{1}{t}=e^{h}-1, \;\ \frac{1}{t}+1=e^{h}\)

\(\displaystyle ln(\frac{1}{t}+1)=h\)

Note that as \(\displaystyle h\to{0}\) then \(\displaystyle t\to{\infty}\)

Make the subs and get \(\displaystyle \lim_{t\to{\infty}}\frac{1}{tln(\frac{1}{t}+1)}\)

\(\displaystyle =\lim_{t\to{\infty}}\frac{1}{ln(\underbrace{1+\frac{1}{t})^{t}}_{\text{e}}}\)

and we get \(\displaystyle \frac{1}{ln(e)}=1\).