Product rule for higher derivatives: Difference between revisions
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| generic point, named functions, point notation || If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>\! \frac{d^n}{dx^n}[f(x)g(x)] = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x)g^{(k)}(x)</math> | | generic point, named functions, point notation || If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>\! \frac{d^n}{dx^n}[f(x)g(x)] = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x)g^{(k)}(x)</math> | ||
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| generic point, named functions, point-free notation || If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math> | | generic point, named functions, point-free notation || If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>(fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}g^{(k)}</math> | ||
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| Pure Leibniz notation || Suppose <math>u</math> and <math>v</math> are both variables functionally dependent on <math>x</math>. Then <math>\frac{d^n(uv)}{(dx)^n} = \sum_{k=0}^n \binom{n}{k} \frac{d^{n-k}u}{(dx)^{n-k}}\frac{d^kv}{(dx)^k}</math> | | Pure Leibniz notation || Suppose <math>u</math> and <math>v</math> are both variables functionally dependent on <math>x</math>. Then <math>\frac{d^n(uv)}{(dx)^n} = \sum_{k=0}^n \binom{n}{k} \frac{d^{n-k}u}{(dx)^{n-k}}\frac{d^kv}{(dx)^k}</math> |
Latest revision as of 06:09, 11 April 2024
ORIGINAL FULL PAGE: Product rule for higher derivatives
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This article is about a differentiation rule, i.e., a rule for differentiating a function expressed in terms of other functions whose derivatives are known.
View other differentiation rules
Statement
Version type | Statement |
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specific point, named functions | This states that if and are times differentiable functions at , then the pointwise product is also times differentiable at , and we have: Here, denotes the derivative of (with , etc.), denotes the derivative of , and is the binomial coefficient. These are the same as the coefficients that appear in the expansion of . |
generic point, named functions, point notation | If and are functions of one variable, the following holds wherever the right side makes sense: |
generic point, named functions, point-free notation | If and are functions of one variable, the following holds wherever the right side makes sense: |
Pure Leibniz notation | Suppose and are both variables functionally dependent on . Then |
One-sided version
There are analogues of each of the statements with one-sided derivatives. Fill this in later
Particular cases
Value of | Formula for |
---|---|
1 | (this is the usual product rule for differentiation). |
2 | . |
3 | . |
4 | |
5 |
Related rules
Similar rules in single variable calculus
- Product rule for differentiation
- Chain rule for higher derivatives
- Chain rule for differentiation
- Repeated differentiation is linear
Similar rules in multivariable calculus
Similar rules in advanced mathematics
Significance
Qualitative and existential significance
Each of the versions has its own qualitative significance:
Version type | Significance |
---|---|
specific point, named functions | This tells us that if and are both times differentiable at a point , so is . |
generic point, named functions, point notation | This tells us that if and are both times differentiable on an open interval, so is . |
generic point, named functions, point-free notation | This shows that the way that behaves is governed by the nature of the derivatives (up to the ) of and . In particular, if and are both continuous functions on an interval, so is . |
Computational feasibility significance
Each of the versions has its own computational feasibility significance:
Version type | Significance |
---|---|
specific point, named functions | This tells us that knowing the values (in the sense of numerical values) of and at a point allows us to compute the value by plugging into the formula and doing a bunch of multiplications and additions. |
generic point, named functions | This tells us that knowledge of the generic expressions for and allows us to compute the generic expression for . |