That is to say, is there a geometrical reason why the derivative of $e^x$ is $e^x$ itself?
Asked
Active
Viewed 469 times
4
-
4Geometrically, the slope of $e^x$ at a point $x$, is precisely the same as the value of the function (i.e. $e^x$). I don't really see more to it than that. – James Aug 11 '14 at 19:17
-
2One way to think about $e^x$ (or, really, (almost) any $a^x$) is that it's partially "self-similar": Stretching it vertically gives you the same curve back, just slid horizontally by an appropriate amount. You can leverage this interpretation to explain the derivative, as in my answer to the question "Could you explain why $\frac{d}{dx}e^x = e^x$ 'intuitively'?" Other answers to that question are insightful, as well. – Blue Aug 11 '14 at 19:40
3 Answers
2
The reason for this depends on your definition of $e$. When you define it as the constant $c$ such that $f'(x)=f(x)$ where $f(x) = c^x$, there is nothing really special about it. For more information, read the Wikipedia page on $e$.
Ragnar
- 6,233
0
The rate of change of the function is the same as the function's value.
Geometrically, if at some $x=a$ the value of the function is $10$, then the slope of the line tangent to the function at $(a,10)$ is also $10$. If at some $x=b$ the value is $500$, then the slope of the line tangent to the function at $(b,500)$ is also $500$.
More generically, $f'(x) = f(x).$
John
- 26,319
0
Note that $$\begin{array}{lll} \frac{d}{dx}a^x&=&\lim_{h\to 0}\frac{a^{x+h}-a^x}{h}\\ &=&\lim_{h\to 0}\frac{a^x(a^{h}-1)}{h}\\ &=&a^x\lim_{h\to 0}\frac{a^{h}-1}{h}\\ &=&a^xf(a) \end{array}$$ Where $$f(a)=\lim_{h\to 0}\frac{a^{h}-1}{h}$$ For some positive values of $a$, $f(a)<1$, and for others, $f(a)>1$. Try to draw some graphs to convince yourself that f(a) is continuous and that it is increasing and concave upwards.
Assuming that the above is true then there is some value of $a$ where $f(a)=1$. We say that if $f(a)=1$ then $a=e$ for some value $e$.
John Joy
- 7,790