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I am trying to prove the derivative of $f(x) = xe^x$ is $f'(x) = e^x + xe^x$ from first principles.

The derivative is $ \frac{f(x+h) - f(x)}{h} $ as h tends to zero

So:

$ \frac{f(x+h) - f(x)}{h} = \frac{(x+h)e^{x+h} - xe^x}{h} $

$ = \frac{x(e^{x+h}-e^x)+he^{x+h}}{h} $

$ = \frac{x(e^{x+h}-e^x)}{h} + e^{x+h} $

This tends to infinity as h tends to zero. Where have I gone wrong?

kabeersvohra
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    How did you determine that the last expression tends to infinity as $h\to 0$? – user26486 Apr 04 '15 at 12:33
  • Then you are having the same problem with differentiating $e^x$... – zoli Apr 04 '15 at 12:33
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    $ \frac{x(e^{x+h}-e^x)}{h} + e^{x+h}=x e^x \frac{e^{h}-1}{h}+ e^{x+h}$ could help. – Claude Leibovici Apr 04 '15 at 12:35
  • @ClaudeLeibovici thanks but then how would I prove $ \frac{e^h-1}{h} $ does not tend to infinity? – kabeersvohra Apr 04 '15 at 12:37
  • Do you know the expression in power series, or Taylor series, for $;e;$ ? – Timbuc Apr 04 '15 at 12:53
  • Just because that, close to $h=0$, $e^h\approx 1+h$. I am sure that you have that somewhere in your notes. Is it OK ? Cheers :-) – Claude Leibovici Apr 04 '15 at 13:04
  • @Timbuc now that's an interesting exercise (no sarcasm intended). How to determine the power series expansion of $e$ with only basic math tools (no fancy tools such as calculating derivatives et al.) – user3371583 Apr 04 '15 at 13:15
  • @user3371583 I don't think that's possible, yet this question's tagged "Differential Equations", so "fancy tools" as derivatives, integrals and the multiplication table of number 8 most likely are old news. – Timbuc Apr 04 '15 at 13:44
  • @Timbuc It's most likely a mistag. Someone who knows differential equations wouldn't be taking derivatives from the definition. – Dylan Apr 05 '15 at 22:44

4 Answers4

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You have incorrectly evaluated the limit of $\frac{e^{x+h}-e^x}{h}$, which tends to $e^x$, rather than to infinity as $h$ tends to zero.

Note that $\frac{e^{x+h}-e^x}{h}$ is the standard difference quotient in the computation of the derivative of $e^x$.

Rolf Hoyer
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  • but surely this equals $ \frac{(e^{x+h})}{h} + \frac{e^x}{h} $ which is infinite regardless because $ \frac{e^x}{h} $ is infinite – kabeersvohra Apr 04 '15 at 12:36
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    how did the minus sign turn into a plus sign? – Rolf Hoyer Apr 04 '15 at 12:37
  • sorry, my bad, how do I prove that $ \frac{e^x+h}{h} $ isnt infinite though? – kabeersvohra Apr 04 '15 at 12:39
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    $\frac{e^{x+h}}{h}$ does tend to infinity, but so too does $\frac{e^x}{h}$. To evaluate their difference is equivalent to taking the derivative of $e^x$. How reasonable this is as a task depends on how you defined $e$. – Rolf Hoyer Apr 04 '15 at 12:43
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Note that $$ \frac{x(e^{x+h}-e^x)}{h}=\frac{x(e^x\cdot e^h-e^h)}{h} = \frac{xe^x(e^h-1)}{h}=xe^x\frac{e^h-1}{h} $$ and the limit of the last factor is the derivative of $t\mapsto e^t$ at $0$, so it is $1$ rather than infinity.

How do you compute it? It depends on how you define $e$: if you define $e$ as the only number $a$ such that $$ \lim_{h\to0}\frac{a^h-1}{h}=1 $$ then you're done.

If you define $e=\lim_{n\to\infty}(1+1/n)^n$ then much more work needs to be done. First show that $$ \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{\!x}=e= \lim_{x\to-\infty}\left(1+\frac{1}{x}\right)^{\!x} $$ Then we can do the substitution $x=1/t$ and get $$ \lim_{t\to0}(1+t)^{1/t}=e $$ so, by continuity of the logarithm, $$ \lim_{t\to0}\frac{\log(1+t)}{t}=1 $$ Now set $\log(1+t)=h$ so that $t=e^h-1$; therefore $$ \lim_{h\to0}\frac{h}{e^h-1}=1 $$

egreg
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If you know/can use power/Taylor series, then

$$\frac{e^h-1}h=\frac{1+h+\mathcal O(h^2)+\ldots-1}{h}=1+\mathcal O(h)\xrightarrow[h\to 0]{}1$$

Timbuc
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Use

$\dfrac{d[f(x)\cdot g(x)]}{dx}=\lim_{h\to0}\dfrac{f(x+h)g(x+h)-f(x)g(x)}h$

$=\lim_{h\to0}f(x+h)\lim_{h\to0}\dfrac{g(x+h)-g(x)}h+g(x)\lim_{h\to0}\dfrac{f(x+h)-f(x)}h$

Set $f(x)=x,g(x)=e^x$ or $g(x)=x,f(x)=e^x$

Finally use $\lim_{h\to0}\dfrac{e^h-1}h=1$

lab bhattacharjee
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  • Im struggling with that last line, how does one prove $ e^h / h $? – kabeersvohra Apr 04 '15 at 12:43
  • @KVohra95, See http://math.stackexchange.com/questions/42679/proof-of-fx-ex-1-x-1-text-as-x-to-0-using-epsilon-delta-definiti and http://math.stackexchange.com/questions/152605/proving-that-lim-limits-x-to-0-fracex-1x-1. There are many other sources available in internet – lab bhattacharjee Apr 04 '15 at 12:45
  • @KVohra95 First prove that $\lim_{y\to 0}\frac{\ln(y+1)}{y}=1$ ($\lim_{y\to 0}\ln\left( (1+y)^{\frac{1}{y}} \right)=\ln\lim_{y\to 0}\left( (1+y)^{\frac{1}{y}} \right)=\ln e = 1$). Now let $u=e^y -1$. Then $\lim_{y\to 0}\frac{e^y-1}{y}=\lim_{u\to 0}\frac{u}{\ln(u+1)}=1$. – user26486 Apr 04 '15 at 12:48