Dx/x of quotient by def of derivative

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The discussion focuses on finding the derivative of the function f(x) = (x^2 - 1) / (2x - 3) using the definition of a derivative rather than the quotient rule. Participants emphasize the importance of expanding the numerator and simplifying the expression to evaluate the limit as h approaches zero. There are also technical notes regarding the use of LaTeX for mathematical expressions, highlighting the need for proper delimiters to avoid rendering issues. The conversation touches on the challenges of previewing LaTeX in forum posts, with suggestions for workarounds. Overall, the thread combines mathematical problem-solving with technical forum usage tips.
karush
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Homework Statement
ok I am retired just doing some review problems
Relevant Equations
definition of derivative
$f(x)=\dfrac{x^2-1}{2x-3}$
ok I just don't see any preview so don't want to add more...
 
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karush said:
Homework Statement: ok I am retired just doing some review problems
Relevant Equations: definition of derivative

$f(x)=\dfrac{x^2-1}{2x-3}$
ok I just don't see any preview so don't want to add more...
You need two dollar signs to delimit your Latex:$$f(x)=\dfrac{x^2-1}{2x-3}$$
 
double \$\$ centers the latex which I don;t want other forums only require one $ wrap $
besides it didm"t render in the preview with double \$\$

anyway this we are supposed to solve by means of def of a derivative which can get quite involved

$$f'( x)=\displaystyle\lim_{h \to 0} \dfrac{{f(x + h)-f(x)}}{h}$$
 
karush said:
anyway this we are supposed to solve by means of def of a derivative which can get quite involved
You need to take the derivative using the original definition, instead of just using the quotient rule for derivitaves?
 
$$f'(x)=\displaystyle\lim_{h \to 0}
\dfrac{\dfrac{(x+h)^2-1}{2(x+h)-3}-\dfrac{x^2-1}{2x-3}}{h}$$
my first inclination is to expand the numerator and combine the fractions
 
karush said:
$$f'(x)=\displaystyle\lim_{h \to 0}
\dfrac{\dfrac{(x+h)^2-1}{2(x+h)-3}-\dfrac{x^2-1}{2x-3}}{h}$$
my first inclination is to expand the numerator and combine the fractions
What else is there to do? Combine, simplify, cancel a factor of ##h## (hopefully) and get a simple limit to evaluate.
 
karush said:
double \$\$ centers the latex which I don;t want other forums only require one $ wrap $
besides it didm"t render in the preview with double \$\$

anyway this we are supposed to solve by means of def of a derivative which can get quite involved

$$f'( x)=\displaystyle\lim_{h \to 0} \dfrac{{f(x + h)-f(x)}}{h}$$
It's two hashes for inline Latex.

You can't preview Latex on the first post. It's just the developers showing who's boss.
 
karush said:
double $$ centers the latex which I don;t want other forums only require one $
One dollar sign is a bad choice for ##\LaTeX## delimiters in forums, because people quite frequently want to say things like "it was $2 for one and $3 for two" and you don't want that to render ##\LaTeX##. So, as others have noted, it's two # for inline maths and two $ for paragraph maths.

The parser here has three oddities, as far as I know because (as much as we love it here) it's a minority-interest add-on to standard forum software. First, it won't render unless there is already ##\LaTeX## on the page - a page refresh cures this. Second, it won't render in preview if there isn't already ##\LaTeX## on the page - a page refresh while in preview mode cures this (the wise poster takes a copy of their text before refreshing). Third, it won't render at all in the new thread preview - I just prepare a thread start as a reply to a random thread that uses ##\LaTeX## and copy-paste into the new thread page when I'm ready.
 

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