Doubt In Explanation of Proof of Chain Rule

In summary, the proof of the Chain Rule in Chapter 3 of Thomas’s Calculus has a limitation in that it does not apply when the function g(x) oscillates rapidly near the origin, leading to a delta u value of 0 even when delta x is not equal to 0. This means that the proof may not apply to any function that is not one-to-one, as delta u will also be 0 in a one-to-one function. An example of a rapidly oscillating function g(x) is y = sin (1/x). However, this limitation only applies when the function is not differentiable at the point of interest. Otherwise, the Chain Rule can still be applied by finding an open interval in which g(x
  • #1
mopit_011
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Homework Statement
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Relevant Equations
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In Chapter 3 of Thomas’s Calculus, they give the following proof of the Chain Rule. After the proof, the text says that this proof doesn’t apply when the function g(x) oscillates rapidly near the origin and therefore leads delta u to be 0 even when delta x is not equal to 0. Doesn’t this proof not apply to any function that isn’t one-to-one them as delta u will be 0 when delta x is not equal to 0 in a one-to-one function?
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  • #2
Could you give us an example of rapidly oscillating function g(x) you said ?
 
  • #3
An example could be a function such as y = sin (1/x).
 
  • #4
Let g(x)= sin 1/x then
[tex]\delta u= \sin \frac{1}{x+\delta x}-\sin \frac{1}{x}[/tex]
Then show me your choice of f(u) also, please.
 
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  • #5
We could let f(u) be a function like u^2.
 
  • #6
mopit_011 said:
An example could be a function such as y = sin (1/x).
This function is poorly behaved near 0 and undefined at 0. It does not have a derivative at 0 and the chain rule will not apply. The chain rule only applies when the functions have derivatives at the points of interest. (see https://en.wikipedia.org/wiki/Chain_rule#Statement)
It is possible to make examples where there is a sequence of points clustering toward the point of interest having ##\Delta u=0## and still having a derivative at the point of interest.
 
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  • #7
mopit_011 said:
Homework Statement:: N/A
Relevant Equations:: N/A

Doesn’t this proof not apply to any function that isn’t one-to-one them as delta u will be 0 when delta x is not equal to 0 in a one-to-one function?
To address the actual question asked, as shown above:
The theorem can apply to most non-one-to-one functions because, provided the function ##g## is differentiable at x and has nonzero derivative there, we can always find an open interval around x in which ##g## is one-to-one. Since we are only aiming to find the derivative of the composed function at x, we can apply the theorem within that open interval to get the result.
For a function that is not differentiable at x, such as g(z) = sin 1/(z - x), we may not be able to find such an open interval where it is one-to-one.
So the key property is being differentiable. Being one-to-one doesn't matter. Accordingly, it's also unnecessary to have a nonzero derivative, as the composed function, if differentiable, will have zero derivative at ##x## if ##g## does,
By the way, ##g(z) = z^3 \sin \frac1z## is a function that is not one-to-one on any interval including 0, but is differentiable at 0, with derivative 0.
 
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  • #8
The OP does not say what specific case the proof is for. Suppose that a function has a sequence of points with equal values that cluster toward the point of interest. If it has a derivative at the point of interest, then the derivative must be zero. In that case, the Chain Rule is true in a practically trivial way. That situation might not be included in the case shown.
 
  • #9
Ref #2-#4
[tex]\triangle u = \sin \frac{1}{x+\triangle x}-\sin \frac{1}{x}[/tex]
[tex]=\sin [\frac{1}{x} (1- \triangle x / x )]-\sin \frac{1}{x} +o(\triangle x^2)[/tex]
[tex]=\sin \frac{1}{x} \cos \triangle x / x^2 - \cos \frac{1}{x} sin \triangle x / x^2 -\sin \frac{1}{x} +o(\triangle x^2)[/tex]
[tex]=( - \frac{1}{x^2}\cos \frac{1}{x})\ \triangle x +o(\triangle x^2)[/tex]
and
[tex]\triangle y = 2u\triangle u + (\triangle u)^2[/tex]
It seems compatible to the proof.

[EDIT]
y=sin 1/x is a good example of applying chain rule, i.e.
[tex]\frac{dy}{dx}=\frac{dy}{d(1/x)} \frac{d (1/x)}{dx} = cos(1/x) (- x^{-2})[/tex]
We have got the same result as above easily.
 
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  • #10
The part of the proof shown stipulates that ##\Delta u \ne 0##. So the case where ##g## is not one-to-one must be handled some other place.
 

FAQ: Doubt In Explanation of Proof of Chain Rule

1. What is the chain rule and why is it important in mathematics?

The chain rule is a fundamental rule in calculus that allows us to find the derivative of a composite function. In other words, it helps us find the rate of change of a function that is made up of multiple smaller functions. This is important in mathematics because many real-world problems involve composite functions, and the chain rule allows us to effectively solve these problems.

2. How do I know when to use the chain rule?

The chain rule is used when we have a function within a function, also known as a composite function. For example, if we have a function f(x) inside another function g(x), then we would use the chain rule to find the derivative of g(f(x)). In general, if we have a function h(x) that can be written as h(g(x)), then we would use the chain rule to find the derivative of h(x).

3. Can you provide an example of how the chain rule is applied?

Sure, let's say we have the function f(x) = (x^2 + 3)^4. We can rewrite this as f(g(x)) where g(x) = x^2 + 3. To find the derivative of f(x), we would use the chain rule, which states that the derivative of f(x) is equal to the derivative of g(x) multiplied by the derivative of the inner function, which in this case is 2x. So, the derivative of f(x) would be 4(x^2 + 3)^3 * 2x = 8x(x^2 + 3)^3.

4. Why do some people doubt the explanation of the proof of the chain rule?

Some people may doubt the explanation of the proof of the chain rule because it can seem abstract and complicated. The proof involves using the limit definition of the derivative and applying it to a composite function, which can be difficult to understand for some people. Additionally, there are different ways to prove the chain rule, so some may doubt the validity of a particular proof.

5. Are there any common mistakes to avoid when using the chain rule?

Yes, there are a few common mistakes to avoid when using the chain rule. One mistake is not correctly identifying the inner and outer functions, which can lead to using the wrong derivative. Another mistake is forgetting to multiply by the derivative of the inner function when using the chain rule. It's important to carefully follow the steps of the chain rule to avoid these errors and ensure an accurate solution.

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