Derivative of 10^x using limit definition

[cmajor47]

Homework Statement

Obtain the first derivative of 10^x by the limit definition.

Homework Equations

f'(x)=lim_h->0 f(x+h)-f(x)/h

The Attempt at a Solution

f'(x)=lim_h->0 10^x+h-10^x/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.

 

[HallsofIvy]

10^x+h= 10^x10^h so
$$\frac{10^{x+h}- 10^x}{h}= 10^x\frac{10^h- 1}{h}$$
Now the question is, what is
$$\lim_{h\rightarrow 0}\frac{10^h- 1}{h}$$?

 

[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?

 

[nrqed]

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.

 

[cmajor47]

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

 

[nrqed]

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!

 

[cmajor47]

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

 

[nrqed]

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

 

[Redbelly98]

... 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.

 

[Redbelly98]

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

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

 

[HallsofIvy]

I suspect this was given as a preliminary to the derivative of e^x so the derivative of e^x cannot be used. It is easy to see that the derivative of a^x, for a any positive number, is a constant times a^x. 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, 10^h= e^{h ln(10)} so
[tex]\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
$$ln(10)\left(\frac{e^{h ln(10)}-1}{h ln(10)}\right)$$
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 10^x is ln(10)10^x.

That is NOT something I would expect a first year calculus student to find for himself!

 

[dynamicsolo]

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.

 

[nrqed]

Moreover, Taylor series are generally taught in second-semester calculus, while covering infinite sequences and series, while the limit

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.

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

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.

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!

 

[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...

 

[RTW69]

10^x=e^(xln10)=e^x, so we are trying to find the derivative of e^x by the limit defination.

The math forum has a solution at:

http://mathforum.org/library/drmath/view/60705.html

 

[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

 

[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