Difference between revisions of "Chain rule for differentiation"
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| significance of uniform bounds on derivatives || If <math>f'</math> and <math>g'</math> are uniformly bounded, then so is <math>(f \circ g)'</math>, with a possible uniform bound being the product of the uniform bounds for <math>f'</math> and <math>g'</math>. | | significance of uniform bounds on derivatives || If <math>f'</math> and <math>g'</math> are uniformly bounded, then so is <math>(f \circ g)'</math>, with a possible uniform bound being the product of the uniform bounds for <math>f'</math> and <math>g'</math>. | ||
+ | |} | ||
+ | |||
+ | ==Examples== | ||
+ | |||
+ | ===Trivial examples=== | ||
+ | |||
+ | We first consider examples where the chain rule for differentiation confirms something we already knew by other means: | ||
+ | |||
+ | {| class="sortable" border="1" | ||
+ | ! Case on <math>f</math> !! Case on <math>g</math> !! What we know about the derivative of <math>f \circ g</math> without using the chain rule !! What the chain rule tells us about the derivative of <math>f \circ g</math> | ||
+ | |- | ||
+ | | a [[constant function]] || any differentiable function || <math>f \circ g</math> is a constant function, so its derivative is the [[zero function]]. || By the chain rule, <math>(f \circ g)'(x) = f'(g(x))g'(x)</math>. <math>f</math> being constant forces <math>f'(g(x))</math> to be zero everywhere, hence the product <math>f'(g(x))g'(x)</math> is also zero everywhere. Thus, <math>(f \circ g)'</math> is also zero everywhere. | ||
+ | |- | ||
+ | | any differentiable function || a [[constant function]] with value <math>k</math> || <math>f \circ g</math> is a constant function with value <math>f(k)</math>, so its derivative is the [[zero function]]. || By the chain rule, <math>(f \circ g)'(x) = f'(g(x))g'(x)</math>. <math>g</math> being constant forces that <math>g'(x) = 0</math> everywhere, hence the product <math>f'(g(x))g'(x)</math> is also zero everywhere. Thus, <math>(f \circ g)'</math> is also zero everywhere. | ||
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+ | | the [[identity function]], i.e., the function <math>x \mapsto x</math> || any differentiable function || <math>f \circ g = g</math>, so <math>(f \circ g)' = g'</math>. || <math>(f \circ g)' = (f' \circ g) \cdot g'</math>. Since <math>f</math> is the function <math>x \mapsto x</matH>, its derivative is the function <math>x \mapsto 1</math>. Plugging this in, we get that <math>f' \circ g</math> is also the constant function <math>x \mapsto 1</math>, so <math>(f \circ g)' = 1g' = g'</math>. | ||
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+ | | any differentiable function || the [[identity function]] || <math>f \circ g = f</math>, so <math>(f \circ g)' = f'</math>. || <math>(f \circ g)' = (f' \circ g) \cdot g'</math>. Since <math>g</math> is the identity function, <math>g'</math> is the function <math>x \mapsto 1</math>. Also, <math>f' \circ g = f'</math>. Thus, <math>(f \circ g)' = f' \cdot 1 = f'</math>. | ||
+ | |- | ||
+ | | the [[square function]] || any differentiable function || <math>f(g(x)) = (g(x))^2</math> and hence its derivative can be computed using the [[product rule for differentiation]]. It comes out as <math>2g(x)g'(x)</math>. || <math>(f \circ g)' = (f' \circ g) \cdot g'</math>. <math>f'</math> is the derivative of the square function, and therefore is <math>x \mapsto 2x</math>. Thus, <math>f'(g(x)) = 2g(x)</math>. We thus get <math>(f \circ g)' = 2g(x)g'(x)</math>. | ||
+ | |} |
Revision as of 00:06, 16 October 2011
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
Contents
Statement for two functions
The chain rule is stated in many versions:
Version type | Statement |
---|---|
specific point, named functions | Suppose and are functions such that is differentiable at a point , and is differentiable at . Then the composite is differentiable at , and we have: |
generic point, named functions, point notation | Suppose and are functions of one variable. Then, we have wherever the right side expression makes sense. |
generic point, named functions, point-free notation | Suppose and are functions of one variable. Then, where the right side expression makes sense, where denotes the pointwise product of functions. |
pure Leibniz notation | Suppose is a function of and is a function of . Then, |
MORE ON THE WAY THIS DEFINITION OR FACT IS PRESENTED: We first present the version that deals with a specific point (typically with a subscript) in the domain of the relevant functions, and then discuss the version that deals with a point that is free to move in the domain, by dropping the subscript. Why do we do this?
The purpose of the specific point version is to emphasize that the point is fixed for the duration of the definition, i.e., it does not move around while we are defining the construct or applying the fact. However, the definition or fact applies not just for a single point but for all points satisfying certain criteria, and thus we can get further interesting perspectives on it by varying the point we are considering. This is the purpose of the second, generic point version.
One-sided version
A one-sided version of sorts holds, but we need to be careful, since we want the direction of differentiability of to be the same as the direction of approach of to . The following are true:
- We require , the function composed first, to be left differentiable at and require to be differentiable at . Then, the left hand derivative of at is times the left hand derivative of at .
- We require , the function composed first, to be right differentiable at and require to be differentiable at . Then, the right hand derivative of at is times the right hand derivative of at .
- We require to be left differentiable at , to be left differentiable at , and we require to be an increasing function. Then, is left differentiable at and the left hand derivative is the left hand derivative of at times the left hand derivative of at .
- We require to be right differentiable at , to be right differentiable at , and we require to be an increasing function. Then, is right differentiable at and the right hand derivative is the right hand derivative of at times the left hand derivative of at .
- We require to be left differentiable at , to be right differentiable at , and we require to be a decreasing function. Then, is left differentiable at and the left hand derivative is the right hand derivative of at times the left hand derivative of at .
- We require to be right differentiable at , to be left differentiable at , and we require to be a decreasing function. Then, is right differentiable at and the right hand derivative is the left hand derivative of at times the left hand derivative of at .
Statement for multiple functions
Suppose are functions. Then, the following is true wherever the right side makes sense:
For instance, in the case , we get:
In point notation, this is:
Related rules
- Chain rule for higher derivatives
- Product rule for differentiation
- Product rule for higher derivatives
- Differentiation is linear
- Inverse function theorem (gives formula for derivative of inverse function).
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 is differentiable at a point and is differentiable at , then is differentiable at . |
generic point, named functions, point notation | If is a differentiable function and is a differentiable function on the intersection of its domain with the range of , then is a differentiable function. |
generic point, named functions, point-free notation | We can deduce properties of based on properties of . In particular, if and are both continuous functions, so is . Another way of putting this is that if and are both continuously differentiable functions, so is . |
Computational feasibility significance
Each of the versions has its own computational feasibility significance:
Version type | Significance |
---|---|
specific point, named functions | If we know the values (in the sense of numerical values) and , we can use these to compute . |
generic point, named functions | This tells us that knowledge of the general expressions for the derivatives of and (along with expressions for the functions themselves) allows us to compute the general expression for the derivative of . |
Computational results significance
Shorthand | Significance |
---|---|
significance of derivative being zero | If , and is differentiable at , then . Note that the conclusion need not follow if is not differentiable at . Also, if and is differentiable at , then . |
significance of sign of derivative | The product of the signs of and gives the sign of . In particular, if both have the same sign, then is positive. If both have opposite signs, then is negative. This is related to the idea that a composite of increasing functions is increasing, and similar ideas. |
significance of uniform bounds on derivatives | If and are uniformly bounded, then so is , with a possible uniform bound being the product of the uniform bounds for and . |
Examples
Trivial examples
We first consider examples where the chain rule for differentiation confirms something we already knew by other means:
Case on | Case on | What we know about the derivative of without using the chain rule | What the chain rule tells us about the derivative of |
---|---|---|---|
a constant function | any differentiable function | is a constant function, so its derivative is the zero function. | By the chain rule, . being constant forces to be zero everywhere, hence the product is also zero everywhere. Thus, is also zero everywhere. |
any differentiable function | a constant function with value | is a constant function with value , so its derivative is the zero function. | By the chain rule, . being constant forces that everywhere, hence the product is also zero everywhere. Thus, is also zero everywhere. |
the identity function, i.e., the function | any differentiable function | , so . | . Since is the function , its derivative is the function . Plugging this in, we get that is also the constant function , so . |
any differentiable function | the identity function | , so . | . Since is the identity function, is the function . Also, . Thus, . |
the square function | any differentiable function | and hence its derivative can be computed using the product rule for differentiation. It comes out as . | . is the derivative of the square function, and therefore is . Thus, . We thus get . |