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mmzaj
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what is quotient rule for higher order derivatives ? i mean the one analogous to http://en.wikipedia.org/wiki/Leibniz_rule_%28generalized_product_rule%29" .
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mmzaj said:i have come to find that
[tex](\frac{f}{g})^{(n)}=\frac{1}{g} \sum^{n}_{k=0} (-1)^k \binom{n+1}{k+1} \frac{(fg^{k})^{(n)}}{g^{k}}[/tex]
but again , this isn't easier to evaluate since we need to expand the term [tex] (fg^{k})^{(n)} [/tex] using both leibniz rule and Faà di Bruno's formula !
mmzaj said:i have come to find that
[tex](\frac{f}{g})^{(n)}=\frac{1}{g} \sum^{n}_{k=0} (-1)^k \binom{n+1}{k+1} \frac{(fg^{k})^{(n)}}{g^{k}}[/tex]
but again , this isn't easier to evaluate since we need to expand the term [tex] (fg^{k})^{(n)} [/tex] using both leibniz rule and Faà di Bruno's formula !
yanshu said:a poor proof: let u = f/g then use leibniz rule tu expand (ug^(k+1))^n,noticing we only need to proove the factor of u^(m)=0 ,if m<n ; the factor of u^(m)=1 ,if m=n. By Faà di Bruno's formula we can get it...
bpet said:It seems true at least up to n=10 checking with Mathematica, but it's not obvious how to derive it. What's the trick?
mmzaj said:i have come to find that
[tex](\frac{f}{g})^{(n)}=\frac{1}{g} \sum^{n}_{k=0} (-1)^k (^{n+1}C_k) \frac{(fg^{k})^{(n)}}{g^{k}}[/tex]
but again , this isn't easier to evaluate since we need to expand the term [tex] (fg^{k})^{(n)} [/tex] using both leibniz rule and Faà di Bruno's formula !
bpet said:...
Moving all terms to the LHS and substituting u=f/g, we need to show that
[tex]\sum_{k=0}^{n+1}(-1)^k (^{n+1}C_k) g^{n+1-k}(g^ku)^{(n)} = 0[/tex]
for all g and u. Expanding the product derivative, we need to show that
[tex]\sum_{k=0}^{n+1}(-1)^k (^{n+1}C_k) g^{n+1-k}(g^k)^{(j)} = 0[/tex]
for all g and all j<=n.
...
The quotient rule for higher order derivatives is a mathematical rule used to find the derivative of a function that is expressed as the ratio of two other functions. It is an extension of the quotient rule for first derivatives, and it is used to find the derivatives of functions with multiple variables.
To apply the quotient rule for higher order derivatives, you first need to find the first and second derivatives of the numerator and denominator. Then, plug those values into the formula: (f'g - g'f) / g^2. This will give you the derivative of the original function.
The purpose of the quotient rule for higher order derivatives is to help find the rate of change of a function that is expressed as a ratio of two other functions. This is useful in many scientific fields, such as physics, economics, and engineering.
Yes, the quotient rule for higher order derivatives can be applied to any function that is expressed as a ratio of two other functions. However, it is important to note that the function must be differentiable for the rule to be applicable.
Yes, some common mistakes when applying the quotient rule for higher order derivatives include incorrectly finding the first and second derivatives of the numerator and denominator, failing to simplify the final answer, and forgetting to use the chain rule when necessary.