Dx/x of quotient by def of derivative

In summary, the derivative of a quotient of two functions can be calculated using the quotient rule. The rule states that if \( u(x) \) and \( v(x) \) are differentiable functions, the derivative of their quotient \( \frac{u}{v} \) is given by \( \frac{u'v - uv'}{v^2} \). This formula allows us to find the rate of change of the quotient by differentiating the numerator and the denominator separately, applying the product of the derivatives accordingly.
  • #1
karush
Gold Member
MHB
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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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  • #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.
 
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FAQ: Dx/x of quotient by def of derivative

What is the definition of the derivative of a quotient?

The derivative of a quotient of two functions, say \( f(x) \) and \( g(x) \), is given by the quotient rule. If \( h(x) = \frac{f(x)}{g(x)} \), then the derivative \( h'(x) \) is \( h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \).

How do you apply the quotient rule in differentiation?

To apply the quotient rule, you need to identify the numerator function \( f(x) \) and the denominator function \( g(x) \). Then, compute the derivatives \( f'(x) \) and \( g'(x) \). Substitute these into the quotient rule formula: \( \left( \frac{f(x)}{g(x)} \right)' = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \).

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 considering the limit definition of the derivative for the function \( h(x) = \frac{f(x)}{g(x)} \) and applying algebraic manipulation and the limit properties, the quotient rule formula can be obtained.

What are some common mistakes to avoid when using the quotient rule?

Common mistakes include: not correctly identifying \( f(x) \) and \( g(x) \), forgetting to square the denominator \( [g(x)]^2 \), incorrectly computing the derivatives \( f'(x) \) and \( g'(x) \), and algebraic errors in simplifying the final expression. Careful attention to each step can help avoid these errors.

When is it better to use the quotient rule instead of simplifying the expression first?

It is often better to use the quotient rule directly when the quotient of the functions is not easily simplified. If the expression can be simplified into a simpler form (like a single polynomial or a product of functions), it may be easier to differentiate using basic rules or the product rule. However, for more complex quotients, the quotient rule is the most straightforward method.

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