7

If $\frac{d}{dx}e{^x} = e{^x}$, then why does $\frac{d}{dx}e^{-14}$ = 0?

Why doesn't $\frac{d}{dx}e^{-14}$ = $e^{-14}$?

I don't understand.

Monica
  • 145
  • 3
    Hint: $e^-14$=$\frac{1}{e^14}$ which is a real number, or a constant. – Guy Corrigall Mar 19 '14 at 00:33
  • 1
    @GuyCorrigall: While I selected an answer, is there any way to denote your response was helpful in understanding the answer? – Monica Mar 19 '14 at 00:46
  • What you really need to remember is that $\frac{d}{dx}(c)=0$ given that $c$ is any and all constant. For instance $c$ could be $2, -98, \pi, \sqrt(2), -e^{-\pi^2}$ etc. –  Mar 19 '14 at 01:07
  • @MitchKnight: Yes, super helpful. It seems like this should be a given, but sometimes it just needs to be spelled out for newbies like me. I actually made notes similar to your comment in my own "study sheets." Again, thanks. Every response adds another element to my understanding. – Monica Mar 19 '14 at 01:15
  • @PedroTamaroff: I don't understand that notation. OR else it is just past my level right now. – Monica Mar 19 '14 at 01:31
  • @PedroTamaroff Actually I find that kind of pedantic writing interesting. Do you have any kind of reference addressing the consistency of taking derivatives with respect to constants? It seems like it would quickly lead to contradictions. – DanielV Mar 19 '14 at 01:32
  • @Monica My point was $$\frac{d}{d\blacktriangle}e^{\blacktriangle}=e^{\blacktriangle}$$ There is a consistency throughout the variables, see? – Pedro Mar 19 '14 at 01:32
  • @Pedro Tamaroff. What you wrote was wrong, you don't differentiate with respect to a constant, just differentiate with respect to the variable and make substitutions later on. –  Mar 19 '14 at 04:47

5 Answers5

11

$e^x$ is a function that depends on $x$. $e^{-14}$ is a constant.

2012ssohn
  • 3,827
3

Because $e^{-14}$ is a constant value. The derivative of any constant value is $0$.

foobar1209
  • 1,047
  • While I selected an answer, is there any way to denote your response was helpful in understanding the answer? I've been finding that it takes several answers to actually help me understand completely. Everyone has their own way of thinking. That you pointed out specifically and simply that the derivative of any constant value is 0 in relation to this problem was excellent. – Monica Mar 19 '14 at 00:47
  • @Monica Outside of upvoting, I don't think so. Either way, it was my pleasure to assist in your understanding. :) – foobar1209 Mar 19 '14 at 00:49
3

If $f(x)=e^x$ there is a difference between take the derivative of $f(x)$ at the point $x=-14$ and derivate $f(-14)$, the first is: $$\lim_{h\to 0}\left(\frac{f(-14+h)-f(-14)}{h}\right) = e^{-14}$$ And the second being a constant function $g(x)=f(-14)$: $$\lim_{h\to 0}\left(\frac{g(-14+h)-g(-14)}{h}\right)=\lim_{h\to 0}\left(\frac{f(-14)-f(-14)}{h}\right) = 0$$

rlartiga
  • 4,205
  • 1
  • 14
  • 24
2

If you look at the graphs of the functions $f(x)=e^x$ and $g(x)=e^{-14}$ then you will notice that the slope of each one is different. The slope at any point $x$ on the function $f(x)=e^x$ is given by $e^x$, since $f'(x)=e^x$. The slope of $g(x)=e^{-14}$ is always zero, since $g(x)=e^{-14}$ for all $x$.

Remember that $e$ is just a number, $e=2.71828 \dots$ and therefore $e^{-14}=(2.7128 \dots)^{-14}=0.00000083152 \dots$

  • While I selected an answer, is there any way to denote your response was helpful in understanding the answer? Your suggestion to graph and visualize it was extremely helpful. I'm really trying to understand these problems on several different levels. – Monica Mar 19 '14 at 00:49
  • To signify that you liked anybody's answer, you click the up arrow to the left of the answer (above the number). Click the down arrow if you don't think it answers your question. –  Mar 19 '14 at 01:00
2

Consider the equation:

$$(\forall x)\, x + 1 > x$$

Obviously in this equation, you can substitute $x = -14$ or whatever you want and get an equally valid equation: $-14 + 1 > -14$.

One the other hand, consider the equation: $$\sum_{x = 1}^4 x + n = 10 + 4n$$

What would it signify to substitute $x=-14$ into the above equation? It would be nonsense. For comparison, what if $n = -14$ were substituted into the above equation? You would get the valid result $\sum_{x = 1}^{4} x + -14 = 1 - 14 + 2 - 14 + 3 - 14 + 4 - 14 = 10 - 4\cdot 14$. That's because, even though we didn't write it, it's understood that the above equation is:

$$(\forall n)\, \sum_{x = 1}^4 x + n = 10 + 4n$$

So you can substitute any $n = \text{ whatever}$ into the above equation because that is what the forall $\forall$ indicates.

Although it's a pain, the equation $$\frac{de^x}{dx} = e^x$$ is not a shorthand for $$(\forall x)\frac{de^x}{dx} = e^x$$ It is more exactly a shorthand for "the derivative of the exponential function is the exponential function", which if you want to see how ugly that actually looks: $$\text{diff}(x \rightarrow e^x) = x \rightarrow e^x$$

That's long and difficult so we don't write that. But the point is that you can't assume an arbitrary substitution like $x = \text{whatever}$ unless the $x$ comes from a $\forall x$.

DanielV
  • 23,556
  • You could have just said it's a constant... Yet you went ahead and posted about some upside down A's. – Shahar Mar 19 '14 at 01:45
  • Not to be rude, but the answer "it's a constant" is wrong. If substitution is valid, then it is valid whether you are substituting for a constant or for a function or for an elephant. Furthermore, a function can be a constant function; distinguishing them in this context is arbitrary. The correct reason is the unwritten binding of the $x$ variable. – DanielV Mar 19 '14 at 02:19