## Derivative of 10^x using limit definition

1. The problem statement, all variables and given/known data
Obtain the first derivative of 10x by the limit definition.

2. Relevant equations
f'(x)=limh->0 f(x+h)-f(x)/h

3. The attempt at a solution
f'(x)=limh->0 10x+h-10x/h
I also know that h=1 as x approaches 0.

Now, how do I make it so that you aren't dividing by h=0.

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 Recognitions: Gold Member Science Advisor Staff Emeritus 10x+h= 10x10h so $$\frac{10^{x+h}- 10^x}{h}= 10^x\frac{10^h- 1}{h}$$ Now the question is, what is $$\lim_{h\rightarrow 0}\frac{10^h- 1}{h}$$?
 To find $$\lim_{h\rightarrow 0}\frac{10^h- 1}{h}$$ I can't plug h=0 in because I would be dividing by 0. Do I plug in 1 since the limit as h approaches 0 is 1?

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## Derivative of 10^x using limit definition

 Quote by cmajor47 To find $$\lim_{h\rightarrow 0}\frac{10^h- 1}{h}$$ I can't plug h=0 in because I would be dividing by 0. Do I plug in 1 since the limit as h approaches 0 is 1?
if you plug h = 0 then 10^h = 1. But you want the next order correction which you explect should be proportional to h. So you expect something of the form

$$lim_{h\rightarrow 0} ~10^h = 1 + a h + \ldots$$

where a is some numerical value and the dots represent terms of higher powers in h. The problem is to find the value of a.

Here is the trick: use that any number x may be written as $$e^{ \ln x}$$. Then use what you know about rules for logs and then Taylor expand the exponential.

 I don't understand what this means. What is "Taylor expanding" and the "next order correction"?

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 Quote by cmajor47 I don't understand what this means. What is "Taylor expanding" and the "next order correction"?
Have you ever seen the relation
$$e^\epsilon = 1 + \epsilon + \frac{\epsilon^2}{2} + \ldots$$ ?

This is a Taylor expansion.

Have you ever proved using the limit definition that the derivative of e^x is e^x? Then you must have used something similar to the above.

If you haven't proved the e^x case and the expansion I wrote above is not familiar to you then I will let someone else help you because I don't see at first sight any other approach.

EDIT: do you know what the limit as h goes to zero of $$\frac{e^h -1}{h}$$ gives? maybe you have been told this without proving it. If you know the result of this limit and are allowed to use it, then I can show you how to finish your problem. If not, I don't see how to help, unfortunately.

Best luck!

 I've never proved e^x. Thank you for trying to help though.

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 Quote by cmajor47 I've never proved e^x. Thank you for trying to help though.
Sorry. I can tell you that the limit as h goes to zero of 10^h is
$$lim_{h \rightarrow 0} 10^h = lim_{h \rightarrow 0} e^{\ln 10^h} = lim_{h \rightarrow 0} e^{h \ln 10} \approx 1 + h \ln 10$$
where I used an identity for logs and then I used the expansion of the exponential I mentioned earlier. Form this you can get the final answer of your question.

Hopefully someone else will be able to find a way to show this result in some other way but I can't think of any!

Best luck

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 Quote by nrqed ... Taylor expand the exponential.
Taylor expansion requires knowledge of what the derivative is. But we don't know the derivative, that is what we are supposed to find.

 Mentor Blog Entries: 10 I don't know if this will be useful, but one might try using the definition of e: $$e = \lim_{N \rightarrow \infty} (1+\frac{1}{N})^N = \lim_{a \rightarrow 0} (1+a)^{1/a}$$ or $$e^A = \lim_{N \rightarrow \infty} (1+\frac{1}{N})^{NA} = \lim_{a \rightarrow 0} (1+a)^{A/a}$$ Also, the fact that $$10^h = e^{h \ln(10)}$$
 Recognitions: Gold Member Science Advisor Staff Emeritus I suspect this was given as a preliminary to the derivative of ex so the derivative of ex cannot be used. It is easy to see that the derivative of ax, for a any positive number, is a constant times ax. It is much harder to determine what that constant is! It's not too difficult to show that, for some values of a, that constant is less than 1 and, for some values of a, larger than 1. Define e to be the number such that that constant is 1. That is, define "e" by $$\lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1$$ As Redbelly98 said, 10h= eh ln(10) so $$\frac{10^h- 1}{h}= \frac{e^{h ln(10)}- 1}{h}[/itex] If we multiply both numerator and denominator of that by ln(10) we get [tex]ln(10)\left(\frac{e^{h ln(10)}-1}{h ln(10)}\right)$$ Clearly, as h goes to 0 so does h ln(10) so if we let k= h ln(10) we have $$ln(10)\left(\lim_{h\rightarrow 0}\frac{e^{h ln(10)}-1}{h ln(10)}\right)= ln(10)\left(\lim_{k\rightarrow 0}\frac{e^k- 1}{k}\right)$$ so the limit is ln(10) and the derivative of 10x is ln(10)10x. That is NOT something I would expect a first year calculus student to find for himself!

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 Quote by Redbelly98 Taylor expansion requires knowledge of what the derivative is. But we don't know the derivative, that is what we are supposed to find.
Moreover, Taylor series are generally taught in second-semester calculus, while covering infinite sequences and series, while the limit

$$\lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1$$

is often demonstrated or proven (if it is not simply stated without proof) in the first-semester course, shortly after having covered limits and while developing the rules of differentiation. I hardly expected that the OP would have seen Taylor series yet...

I believe the proof given in textbooks usually revolves around the limit definition of e, which Redbelly98 gives in post #10.

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 $$\lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1$$ is often demonstrated or proven (if it is not simply stated without proof) in the first-semester course, shortly after having covered limits and while developing the rules of differentiation. I hardly expected that the OP would have seen Taylor series yet... I believe the proof given in textbooks usually revolves around the limit definition of e, which Redbelly98 gives in post #10.
 Recognitions: Homework Help I suspect that OP's textbook presents the limit $$\lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1$$ somewhere in the chapter and that a student is just asked to recognize that they could apply it, in something like the manner Halls suggests in post #11...