# Computing the derivative of an exponential function

• member 731016
member 731016
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

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.

member 731016
FactChecker said:
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.

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

Many thanks!

Callumnc1 said:
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##.

member 731016
Callumnc1 said:
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## .

malawi_glenn, member 731016 and Mark44
FactChecker said:
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##.

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!

malawi_glenn
SammyS said:
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## .

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

Many thanks!

malawi_glenn
Callumnc1 said:

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:
Callumnc1 said:
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.

Last edited:
member 731016 and malawi_glenn
Callumnc1 said:

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.

member 731016
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?

Last edited:
member 731016

## What is the derivative of the exponential function $$e^x$$?

The derivative of the exponential function $$e^x$$ is $$e^x$$. This is because the exponential function $$e^x$$ is unique in that it is its own derivative.

## How do you find the derivative of $$a^x$$, where $$a$$ is a constant?

To find the derivative of $$a^x$$, where $$a$$ is a constant, you use the formula $$\frac{d}{dx} a^x = a^x \ln(a)$$. This incorporates the natural logarithm of the base $$a$$.

## What is the derivative of $$e^{u(x)}$$, where $$u(x)$$ is a function of $$x$$?

The derivative of $$e^{u(x)}$$ is found using the chain rule. It is $$\frac{d}{dx} e^{u(x)} = e^{u(x)} \cdot u'(x)$$, where $$u'(x)$$ is the derivative of $$u(x)$$.

## How do you compute the derivative of $$a^{u(x)}$$, where $$a$$ is a constant and $$u(x)$$ is a function of $$x$$?

To compute the derivative of $$a^{u(x)}$$, you use the chain rule combined with the formula for the derivative of $$a^x$$. The result is $$\frac{d}{dx} a^{u(x)} = a^{u(x)} \ln(a) \cdot u'(x)$$, where $$u'(x)$$ is the derivative of $$u(x)$$.

## Why is the natural logarithm used in the derivatives of exponential functions with bases other than $$e$$?

The natural logarithm is used because it simplifies the differentiation process. When differentiating $$a^x$$, the natural logarithm $$\ln(a)$$ appears naturally from the chain rule applied to the exponential function and the properties of logarithms, making it a crucial component in the derivative formula.

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