# Modulus Sin(pi X ) issue.

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#1
why is |sin(pi(x)))| not differentiable at multiples of pi?
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5 years ago
#2
why is |sin(pi(x)))| not differentiable at multiples of pi?
Do you mean |sin x|?
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#3
(Original post by morgan8002)
Do you mean |sin x|?
well yes but why are pi multiples of |sin(x) | not differentiable?

Such as |sin(pi*x)|
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5 years ago
#4
why is |sin(pi(x)))| not differentiable at multiples of pi?
You mean when . Go here and graph the function:

https://www.desmos.com/calculator

The answer should then be clear.
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#5
(Original post by atsruser)
You mean when . Go here and graph the function:

https://www.desmos.com/calculator

The answer should then be clear.
Ah, so is it because there is sharp turn when it meets the x axis which would mean non differentiability at those points?

edit: Which would mean it is non differential for all modulus functions?
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5 years ago
#6
Ah, so is it because there is sharp turn when it meets the x axis which would mean non differentiability at those points?
Right. The problem is that the slope when you approach those points from the left differs from that when you approach them from the right. So there is no unique slope. The limits have to be the same from both directions for the derivative to exist.
1
5 years ago
#7
Ah, so is it because there is sharp turn when it meets the x axis which would mean non differentiability at those points?

edit: Which would mean it is non differential for all modulus functions?
No. for all is a counter-example.
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#8
(Original post by atsruser)
No. for all is a counter-example.
Ah silly of me. Thank you.
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5 years ago
#9
(Original post by atsruser)
No. for all is a counter-example.
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5 years ago
#10
(Original post by morgan8002)
Well, which is differentiable everywhere. Or am I missing your point?
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5 years ago
#11
edit: Which would mean it is non differential for all modulus functions?
There are lots of more interesting functions of the form of course. For example, so that is fine, as is - essentially anything that is always +ve or -ve over its domain will be fine, as long as the non-modulused function is differentiable. (... waits for some obvious counter-example that I've missed ...)
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5 years ago
#12
(Original post by atsruser)
There are lots of more interesting functions of the form of course. For example, so that is fine, as is - essentially anything that is always +ve or -ve over its domain will be fine, as long as the non-modulused function is differentiable. (... waits for some obvious counter-example that I've missed ...)
Probably just me missing something, but why don't we run into problems when we use the chain rule on |x^2| at 0? Clearly it exists at 0 since the function is the same as x^2 but doing it fully requires you to evaluate the derivative of |x| at 0 which is undefined??
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5 years ago
#13
Probably just me missing something, but why don't we run into problems when we use the chain rule on |x^2| at 0? Clearly it exists at 0 since the function is the same as x^2 but doing it fully requires you to evaluate the derivative of |x| at 0 which is undefined??
The range of is so when you differentiate the composite function , the modulus function in question is a restriction to a smaller domain i.e. you are working with

So when we differentiate it using the chain rule, we have no way of breaking the modulus part by approaching 0 from both +ve and -ve numbers - we only need to deal with a one-sided limit.
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5 years ago
#14
Probably just me missing something, but why don't we run into problems when we use the chain rule on |x^2| at 0? Clearly it exists at 0 since the function is the same as x^2 but doing it fully requires you to evaluate the derivative of |x| at 0 which is undefined??
Well, we can always pull out the easy answer that the chain rule requires the composite derivatives to exist... but in this case we can 'cheat' a bit because, if f(x)=|x^2|:=|u|, then |u|=u because u is greater than or equal to zero on the domain (real x). So d/du |u|=d/du u, which exists.
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5 years ago
#15
(Original post by Implication)
Well, we can always pull out the easy answer that the chain rule requires the composite derivatives to exist... but in this case we can 'cheat' a bit because, if f(x)=|x^2|:=|u|, then |u|=u because u is greater than or equal to zero on the domain (real x). So d/du |u|=d/du u, which exists.
Isn't this just the question though? I know it exists, I want to know why doing it 'properly' doesn't work

(Original post by atsruser)
The domain of is so when you differentiate the composite function , the modulus function in question is a restriction to a smaller domain i.e. you are working with

So when we differentiate it using the chain rule, we have no way of breaking the modulus part by approaching 0 from both +ve and -ve numbers - we only need to deal with a one-sided limit.
Sorry, did you mean range? I don't see why negative numbers shouldn't be a valid input for x^2
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5 years ago
#16
Isn't this just the question though? I know it exists, I want to know why doing it 'properly' doesn't work
What do you mean by 'properly'? The chain rule goes something like:

suppose:
f(x)=f(u(x))
df/dg exists at g(c)
dg/dx exists at c

then:
at c, df/dx=df/dg * dg/dx

if one of the derivatives does not exist, the chain rule is not valid.
basically, the chain rule says that A implies B
but we don't have A
so we can't conclude B

As it so happens in the example you gave (and as atsruser explained above better than i did), the domain of the modulus in this case is the positive numbers, for which the derivative does exist. So in this case we can use the chain rule.

Sorry, did you mean range? I don't see why negative numbers shouldn't be a valid input for x^2
I believe he meant range of x^2, which would be the domain of f(u) for u=x^2
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5 years ago
#17
(Original post by Implication)
What do you mean by 'properly'? The chain rule goes something like:

suppose:
f(x)=f(u(x))
df/dg exists at g(c)
dg/dx exists at c

then:
at c, df/dx=df/dg * dg/dx

if one of the derivatives does not exist, the chain rule is not valid.
basically, the chain rule says that A implies B
but we don't have A
so we can't conclude B

As it so happens in the example you gave (and as atsruser explained above better than i did), the domain of the modulus in this case is the positive numbers, for which the derivative does exist. So in this case we can use the chain rule.

I believe he meant range of x^2, which would be the domain of f(u) for u=x^2
But wouldn't the chain rule in this case require evaluating the derivative of |x| at 0 (since 0^2 is still 0)? I get why this failing doesn't say anything about the limit existing thanks to your post, but how can you get around d/dg |g| at 0?

Thanks, that makes more sense
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5 years ago
#18
In order for a function to be differentiable at a point, the limit needs to be the same from both sides.

It's the same reason for why you can't define the gradient of 1/x at 0.
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5 years ago
#19
(Original post by atsruser)
Well, which is differentiable everywhere. Or am I missing your point?
Yes. I see the miscommunication now. I thought you meant the single counterexample f(x) = |x|, defined on , while you actually meant the family of counterexamples f(x) = |c| for some fixed real c.
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5 years ago
#20
Sorry, did you mean range? I don't see why negative numbers shouldn't be a valid input for x^2
Yes, I meant range. The range of becomes the domain of
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