Factoring 'h' out in difference quotient

[wizkhal]

Hi all, I'm a beginner in calculus so my question might be stupid. When a function is differentiable, then in difference quotient one can always factor 'h' out in the numerator, even if the function is exponential and 'h' is in the exponent. Is some magic behind this or something else?
I've read lot of math-texts to understand this, but all I've found is a text at "www.karlscalculus.org":
"(...) f(x + h) always expands to: f(x + h) = f(x) + Δf(x, h)
(...) Whatever content Δf(x, h) has in addition to h f'(x) must shrink away to insignificance when compared with h f'(x). Why? Because if it didn't, the limit would not go to f'(x). "
Is there some better way to explain this?
Thanks in advance.
W

 

[Stephen Tashi]

The answer is no, in general. The special case when f(x) is a polynomial, you can. (Well, I suppose f(x) =4 is technically a polynomial and there is no h to factor, but if the polynomial is not constant, you can.)

For example, look at f(x) = 5^x

f(x+h) - f(x) = 5^{x+h} - 5^x = 5^x ( 5^h - 1)

The difference quotient is 5^x \frac{( 5^h -1)}{h}

When symbolic manipulations fail, someone must resort to actual thinking. Finding the derivative involves evaluating

\lim_{h \rightarrow 0} \frac{5^h - 1}{h}

Your textbook probably discusses such limits and the special case where the number is e instead of 5.
In that case the limit is 1, so the derivative of e^x [/tex] is (e^x)(1) = e^x.<br /> <br /> This should remind you that you cannot apply the rule &quot;multiply by the exponent and subtract one from the exponent&quot; when you take the derivative of functions where the variable is in the exponent itself.<br /> <br /> The differentials for functions like sin(x), cos(x) etc. also don&#039;t permit factoring out an h.

 

[wizkhal]

Thank you for the answer.

For example, look at f(x) = 5^x

f(x+h) - f(x) = 5^{x+h} - 5^x = 5^x ( 5^h - 1)

The difference quotient is 5^x \frac{( 5^h -1)}{h}

When symbolic manipulations fail, someone must resort to actual thinking. Finding the derivative involves evaluating

\lim_{h \rightarrow 0} \frac{5^h - 1}{h}

Can't I continue with this difference quotient in such a way (with factoring 'h' out) ? :

\lim_{h \rightarrow 0} \frac{5^h - 1}{h} = \lim_{h \rightarrow 0} \frac{e^{hln5} - 1}{h} = \lim_{h \rightarrow 0} \frac{(1 + hln5) - 1}{h} = \lim_{h \rightarrow 0} \frac{hln5}{h} = ln5

 

[gb7nash]

Can't I continue with this difference quotient in such a way (with factoring 'h' out) ? :

\lim_{h \rightarrow 0} \frac{e^{hln5} - 1}{h} = \lim_{h \rightarrow 0} \frac{(1 + hln5) - 1}{h}

How is this true?

 

[Stephen Tashi]

Can't I continue with this difference quotient in such a way (with factoring 'h' out) ? :

\lim_{h \rightarrow 0} \frac{5^h - 1}{h} = \lim_{h \rightarrow 0} \frac{e^{hln5} - 1}{h} = \lim_{h \rightarrow 0} \frac{(1 + hln5) - 1}{h} = \lim_{h \rightarrow 0} \frac{hln5}{h} = ln5

You apparently rely on the approximation the McLaurin series e^q = 1/0! + q/1! + q^2/2! + ... [/tex] which is approximately 1 + q for small values of q.. So you left out an infinite number of terms.<br /> <br /> If we assume f(x) has a Taylor series expansion then you can write<br /> f(x+h) - f(x) = ( f(x) + \frac{ h f&amp;#039;(x)}{1!} + ...) - f(x) <br /> which involves an infinite series in h. You might develop some mathematical theorems that describe when you can &quot;factor out&quot; an h out of an infinite series, but this is not &quot;factoring out&quot; as is done in ordinary algebra.

 

[gb7nash]

You apparently rely on the approximation the McLaurin series e^q = 1/0! + q/1! + q^2/2! + ... [/tex] which is approximately 1 + q for small values of q.. So you left out an infinite number of terms.<br /> <br /> If we assume f(x) has a Taylor series expansion then you can write<br /> f(x+h) - f(x) = ( f(x) + \frac{ (x-h)f&amp;#039;(x)}{1!} + ...) - f(x) <br /> which involves an infinite series in h. You might develop some mathematical theorems that describe when you can &quot;factor out&quot; an h out of an infinite series, but this is not &quot;factoring out&quot; as is done in ordinary algebra.

<br /> <br /> To add to this, this also makes use of a derivative, which we&#039;re trying to find! This process is circular and not a good idea.

 

[HallsofIvy]

Thank you for the answer.



Can't I continue with this difference quotient in such a way (with factoring 'h' out) ? :

\lim_{h \rightarrow 0} \frac{5^h - 1}{h} = \lim_{h \rightarrow 0} \frac{e^{hln5} - 1}{h} = \lim_{h \rightarrow 0} \frac{(1 + hln5) - 1}{h} = \lim_{h \rightarrow 0} \frac{hln5}{h} = ln5

Just to make very clear what others have said, e^{5ln(x)} is NOT equal to "1+ hln(5)". That is a linear approximation for which you need either the Taylor's expansion of e^x or, at least, the tangent approximation to e^x, both of which require that you know the derivative in advance.

 

[Hurkyl]

It does depend on how generous you want to be with "factor out".