Computing the derivative of an exponential function

[ChiralSuperfields]

Homework Statement

I am trying to understand why as h approach's zero $\frac{b^h - 1}{h} = f'(0)$. Dose anybody please know of a good way to explain this? Many thanks!

Relevant Equations

Pls see below

 

[FactChecker]

That is the definition of the derivative of $f(x) = b^x$ at $x=0$. Notice that they do not say there that the limit and the derivative exist. In fact, the last sentence says "if the exponential function $f(x) = b^x$ is differentiable" (emphasis mine). So there is still something to prove.

 

[ChiralSuperfields]

That is the definition of the derivative of $f(x) = b^x$ at $x=0$. Notice that they do not say there that the limit and the derivative exist. In fact, the last sentence says "if the exponential function $f(x) = b^x$ is differentiable" (emphasis mine). So there is still something to prove.

Thank you for your reply @FactChecker !

How dose $\lim_{x \rightarrow h} {\frac {b^h - 1} {h}} = f'(0)$?

Many thanks!

 

[FactChecker]

How dose $\lim_{x \rightarrow h} {\frac {b^h - 1} {h}} = f'(0)$?

You have formula wrong. I think that $\lim_{h \rightarrow 0} {\frac {f( x+h) - f(x)}{h}} = f'(x)$ is a common definition of the derivative. Is that your definition of the derivative? Now plug in $x=0$.

 

[SammyS]

Homework Statement:: I am trying to understand why as h approach's zero $\frac{b^h - 1}{h} = f'(0)$. Dose anybody please know of a good way to explain this? Many thanks!
Relevant Equations:: Pls see below

First: The word is "Please", not Pls .

Also, you've been misspelling the word "Does". It is not "Dose".

So, you're back to posting a very detailed explanation of some math or physics topic, then asking some question regarding a detail which has been very well explained.

The best answer I can give as to why $\displaystyle f'(0)=\lim_{h\to 0} \dfrac{b^h - 1}{h}$ is because $\displaystyle b^0 = 1$ .

 

[ChiralSuperfields]

You have formula wrong. I think that $\lim_{h \rightarrow 0} {\frac {f( x+h) - f(x)}{h}} = f'(x)$ is a common definition of the derivative. Is that your definition of the derivative? Now plug in $x=0$.

Thank you for your reply @FactChecker!

If I plug in x = 0, to the definition of derivative (the expression that you mentioned, I get

$\frac{0}{0} = undefined$. I am not sure where to go from here?

Many thanks!

 

[ChiralSuperfields]

First: The word is "Please", not Pls .

Also, you've been misspelling the word "Does". It is not "Dose".

So, you're back to posting a very detailed explanation of some math or physics topic, then asking some question regarding a detail which has been very well explained.

The best answer I can give as to why $\displaystyle f'(0)=\lim_{h\to 0} \dfrac{b^h - 1}{h}$ is because $\displaystyle b^0 = 1$ .

Thank you for your reply @SammyS!

But plugging in h = 0 gives $\frac{0}{0} = undefined$. How can $f'(0)$ be undefined since it is not a number?

Many thanks!

 

[FactChecker]

Thank you for your reply @FactChecker!

If I plug in x = 0, to the definition of derivative (the expression that you mentioned, I get

$\frac{0}{0} = undefined$. I am not sure where to go from here?

Many thanks!

Actually, you went too far. You asked this:

Homework Statement:: I am trying to understand why as h approach's zero $\frac{b^h - 1}{h} = f'(0)$.

Plug in $x=0$ to get the definition $f'(0) =\lim_{h \rightarrow 0} \frac{b^{0+h} - b^0}{h} = \lim_{h \rightarrow 0}\frac {b^h - 1}{h}$.
Notice that this is the definition, it does not say that the limits actually exist or the value of the limit. You are trying to prove that, which is a step too far for now. I assume that is proven somewhere else.

 

[SammyS]

Thank you for your reply @SammyS!

But plugging in h = 0 gives $\frac{0}{0} = undefined$. How can $f'(0)$ be undefined since it is not a number?

Many thanks!

That's not how to evaluate this limit.

You are reading something into what is stated which is not there.

 

[malawi_glenn]

h should not be zero, should only be close to zero.

Try this, set b =2.72 and evalute that fraction at h = 0.001 and for h = 10^-6 and for h = 10^-9

You should see a pattern now.

Repeat the above but for b = 2.7183

Wha do you find?

 

[ChiralSuperfields]

Sorry completely forgot about this thread. I will come back here.