Dx/x of quotient by def of derivative

  • #1
karush
Gold Member
MHB
3,269
5
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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  • #2
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:$$f(x)=\dfrac{x^2-1}{2x-3}$$
 
  • #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

$$f'( x)=\displaystyle\lim_{h \to 0} \dfrac{{f(x + h)-f(x)}}{h}$$
 
  • #4
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?
 
  • #5
$$f'(x)=\displaystyle\lim_{h \to 0}
\dfrac{\dfrac{(x+h)^2-1}{2(x+h)-3}-\dfrac{x^2-1}{2x-3}}{h}$$
my first inclination is to expand the numerator and combine the fractions
 
  • #6
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.
 
  • #7
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.
 
  • #8
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.
 
  • Informative
Likes PeroK

1. What is the definition of the derivative of a quotient?

The derivative of a quotient of two functions f(x) and g(x) is given by the formula (f'(x)g(x) - f(x)g'(x)) / (g(x))^2.

2. How do you find the derivative of a quotient using the definition?

To find the derivative of a quotient using the definition, you first differentiate the numerator and denominator separately, then apply the quotient rule formula mentioned in the previous answer.

3. Why is the quotient rule used to find the derivative of a quotient?

The quotient rule is used because the derivative of a quotient of two functions is not simply the quotient of their derivatives. The quotient rule takes into account the relationship between the functions in the numerator and denominator.

4. Can the quotient rule be derived from the definition of the derivative?

Yes, the quotient rule can be derived from the definition of the derivative by applying the limit definition of the derivative to the ratio of two functions and simplifying the result.

5. How is the derivative of a quotient related to the chain rule?

The derivative of a quotient is related to the chain rule in cases where the numerator and denominator are composed of functions of x. In such cases, the chain rule is applied to differentiate the composite functions before using the quotient rule.

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