# Derivative of 10^x using limit definition

1. Jun 29, 2008

### cmajor47

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.

2. Jun 29, 2008

### HallsofIvy

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
$$\lim_{h\rightarrow 0}\frac{10^h- 1}{h}$$?

3. Jun 29, 2008

### 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?

4. Jun 29, 2008

### nrqed

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.

5. Jun 29, 2008

### cmajor47

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

6. Jun 29, 2008

### nrqed

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!

7. Jun 29, 2008

### cmajor47

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

8. Jun 29, 2008

### nrqed

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

9. Jun 29, 2008

### Redbelly98

Staff Emeritus
Taylor expansion requires knowledge of what the derivative is. But we don't know the derivative, that is what we are supposed to find.

10. Jun 29, 2008

### Redbelly98

Staff Emeritus
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

$$10^h = e^{h \ln(10)}$$

11. Jun 29, 2008

### HallsofIvy

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!

12. Jun 29, 2008

### dynamicsolo

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.

13. Jun 29, 2008

### nrqed

Agreed. I should not have mentioned Taylor series. They seem so natural to me now that I tend to use them without even thinking about it.
This is why I then asked the OP if he/she had seen the formula you wrote just above. I hope he/she has. Because if he/she has to go back to the limit definition and prove the above identity in order to solve the original question, this problem seems much more challenging than I would expect as an assignment problem at that level!!

14. Jun 29, 2008

### dynamicsolo

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...

15. Jun 30, 2008

### RTW69

16. Jun 30, 2008

### RTW69

Sorry, the dereivative of 10^x is of course ln10(10^x) but the Math forum derivation of the derivative of e^x is still helpful

17. Jun 30, 2008

### RTW69

More thoughts on this problem.

f(x)=10^X=e^(x*ln10)

f(x+h)= e^(ln10(x+h))=e^(ln10*x)*e^(ln10*h)

Plugging into definitation of derivative and simplifying gives

f'(x)= limit(h goes to 0) 10^x(10^h-1)/h

tabulating the limit as h goes to 0 of (10^h-1)/h= ln10