# question about derivative of exponential function

When I proved derivation the exponential function expose with problem that have to use derivative of $e^x$ $$\frac{de^x}{dx} = \lim_{h\to 0}\frac{e^{x+h} -e^x}h=\lim_{h\to 0} e^x \frac{e^h-1}h =e^x \cdot \lim_{h\to 0} \frac{e^h-1}h$$

Calculate $\displaystyle\lim_{h\to 0} \frac{e^h-1}h$ but can’t use l’hopital theorem and Taylors theorem because use derivative of $e^x$ . Please help me to solve it.

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How was $e^h$ defined? –  Berci Nov 13 '12 at 13:08

I assume, you can use that $\ e^h=\displaystyle\lim_{n\to\infty}\left(1+\frac hn \right)^n$.

Hint: Use the Bernoulli inequation: $(1+x)^{\alpha} \ge 1+\alpha x$ if $x > -1$ and $\alpha>0$, so it yields $e^h\ge 1+h\$ if $\ h> -1$, and take its reciprocal for the converse to prove that the limit you look for is $1$.

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With your inequality I can prove that $\lim (e^h-1)/h \geq 1$. How do we prove equality? –  An old man in the sea. Oct 17 '14 at 13:07

Sometimes the fact that that limit is $1$ is taken to be the definition of $e$. (That's what's done in Stewart's textbook.)

Supposing you had $\dfrac{d}{dx} 4^x$ instead of $\dfrac{d}{dx} e^x$. You'd get $4^x\cdot\lim\limits_{h\to0}\dfrac{4^h - 1}{h}$.

But the function $y=4^x$ gets steeper as you go from left to right, and the slope of its secant line through the points where $x=-1/2$ and $x=0$ is $1$; therefore the slope of the curve at $x=0$ is more than $1$. If you had $y=2^x$, you'd consider the secant line at $x=0$ and $x=1$ and conclude that the slope at $0$ is less than $1$.

So $4$ is too big, and $2$ is too small, to be the base of the natural exponential function. Somewhere between $2$ and $4$ is the right number. If $e$ is that right number, then of course the slope at $0$ is $1$.

(What this omits is how we know just where that number is, i.e. that it's $2.71828\ldots$, beyond the fact that it's between $2$ and $4$.)

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Use the definition of limit: $$\lim _{ h\rightarrow 0 }{ \frac { { e }^{ x+h }-{ e }^{ x } }{ h } } ={ e }^{ x }\lim _{ h\rightarrow 0 }{ \frac { { e }^{ h }-1 }{ h } } ={ e }^{ x }\lim _{ h\rightarrow 0 }{ \frac { { e }^{ h }-{ e }^{ 0 } }{ h } } .$$

$\lim _{ h\rightarrow 0 }{ \frac { { e }^{ h }-{ e }^{ 0 } }{ h } }$ is the derivative of $e^x$ at $x=0$. Therefore we can write $\frac { d }{ dx } { e }^{ x }$ as $$\frac { d }{ dx } { e }^{ x }={ e }^{ x }{ \left[ \frac { d }{ dx } { e }^{ x } \right] }_{ x=0 }.$$ Open ${ \left[ \frac { d }{ dx } { e }^{ x } \right] }_{ x=0 }$ in a similar way, you get $${ \left[ \frac { d }{ dx } { e }^{ x } \right] }_{ x=0 }={ e }^{ 0 }{ \left[ \frac { d }{ dx } { e }^{ x } \right] }_{ x=0 }={ e }^{ 0 }{ e }^{ 0 }{ \left[ \frac { d }{ dx } { e }^{ x } \right] }_{ x=0 }=1\cdot 1\cdot 1\cdot 1\cdot 1\cdot 1\cdots .$$ As a result $$\frac { d }{ dx } { e }^{ x }={ e }^{ x }.$$

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The last part, even if correct doesn't seem to prove what you want... –  An old man in the sea. Oct 17 '14 at 12:50