# Modulus Sin(pi X ) issue.

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#2

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why is |sin(pi(x)))| not differentiable at multiples of pi?

**Shady778**)why is |sin(pi(x)))| not differentiable at multiples of pi?

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(Original post by

Do you mean |sin x|?

**morgan8002**)Do you mean |sin x|?

Such as |sin(pi*x)|

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#4

(Original post by

why is |sin(pi(x)))| not differentiable at multiples of pi?

**Shady778**)why is |sin(pi(x)))| not differentiable at multiples of pi?

https://www.desmos.com/calculator

The answer should then be clear.

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(Original post by

You mean when . Go here and graph the function:

https://www.desmos.com/calculator

The answer should then be clear.

**atsruser**)You mean when . Go here and graph the function:

https://www.desmos.com/calculator

The answer should then be clear.

edit: Which would mean it is non differential for all modulus functions?

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#6

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Ah, so is it because there is sharp turn when it meets the x axis which would mean non differentiability at those points?

**Shady778**)Ah, so is it because there is sharp turn when it meets the x axis which would mean non differentiability at those points?

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#7

(Original post by

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?

**Shady778**)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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#11

(Original post by

edit: Which would mean it is non differential for all modulus functions?

**Shady778**)edit: Which would mean it is non differential for all modulus functions?

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#12

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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 ...)

**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 ...)

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#13

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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??

**silentshadows**)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??

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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#14

**silentshadows**)

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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#15

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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.

**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.

(Original post by

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.

**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.

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#16

(Original post by

Isn't this just the question though? I know it exists, I want to know why doing it 'properly' doesn't work

**silentshadows**)Isn't this just the question though? I know it exists, I want to know why doing it 'properly' doesn't work

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

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#17

(Original post by

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

I believe he meant range of x^2, which would be the domain of f(u) for u=x^2

**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

Thanks, that makes more sense

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#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.

It's the same reason for why you can't define the gradient of 1/x at 0.

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#19

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#20

(Original post by

Sorry, did you mean range? I don't see why negative numbers shouldn't be a valid input for x^2

**silentshadows**)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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