Trouble checking whether a function is differentiable or not
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I have this function $x^2|cos frac pi x|$ . To check if it's differentiable at $x=0$, I first directly differentiated the given function- having two cases, opening the mod sign with positive and negative signs. In both cases I end up with a $sin frac pi x $ term in the derivative, so that at $x=0$ the derivative is discontinuous (oscillatory discontinuity) and so the derivative shouldn't exist at $x=0$. But, if I differentiate using first principles instead, I get the derivative as $0$, and so the function IS differentiable at $x=0$. So, is the function differentiable or not, which result do I take?
calculus
edited Jan 5 at 13:13
Kenny Wong
18.3k21438
asked Jan 5 at 13:09
karun mathewskarun mathews
245
$endgroup$
|
show 3 more comments
$begingroup$
I have this function $x^2|cos frac pi x|$ . To check if it's differentiable at $x=0$, I first directly differentiated the given function- having two cases, opening the mod sign with positive and negative signs. In both cases I end up with a $sin frac pi x $ term in the derivative, so that at $x=0$ the derivative is discontinuous (oscillatory discontinuity) and so the derivative shouldn't exist at $x=0$. But, if I differentiate using first principles instead, I get the derivative as $0$, and so the function IS differentiable at $x=0$. So, is the function differentiable or not, which result do I take?
calculus
edited Jan 5 at 13:13
Kenny Wong
18.3k21438
asked Jan 5 at 13:09
karun mathewskarun mathews
245
$endgroup$
2
$begingroup$
You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
$endgroup$
– metamorphy
Jan 5 at 13:14
1
$begingroup$
To be clear, "differentiable" means "the derivative exists" (which you showed from first principles). It does not mean "the derivative is itself continuous" (which you showed is false here).
$endgroup$
– Mark S.
Jan 5 at 14:05
$begingroup$
But I don't understand, if by differentiation using first principles, I could find the left hand and right hand limits of the derivative, then why is the function still discontinuous? Is it because we can't find the value of the derivative AT x=0?
$endgroup$
– karun mathews
Jan 5 at 14:43
$begingroup$
Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
$endgroup$
– karun mathews
Jan 5 at 14:48
1
$begingroup$
@karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
$endgroup$
– Matt Samuel
Jan 7 at 14:35
|
show 3 more comments
$begingroup$
I have this function $x^2|cos frac pi x|$ . To check if it's differentiable at $x=0$, I first directly differentiated the given function- having two cases, opening the mod sign with positive and negative signs. In both cases I end up with a $sin frac pi x $ term in the derivative, so that at $x=0$ the derivative is discontinuous (oscillatory discontinuity) and so the derivative shouldn't exist at $x=0$. But, if I differentiate using first principles instead, I get the derivative as $0$, and so the function IS differentiable at $x=0$. So, is the function differentiable or not, which result do I take?
calculus
edited Jan 5 at 13:13
Kenny Wong
18.3k21438
asked Jan 5 at 13:09
karun mathewskarun mathews
245
$endgroup$
I have this function $x^2|cos frac pi x|$ . To check if it's differentiable at $x=0$, I first directly differentiated the given function- having two cases, opening the mod sign with positive and negative signs. In both cases I end up with a $sin frac pi x $ term in the derivative, so that at $x=0$ the derivative is discontinuous (oscillatory discontinuity) and so the derivative shouldn't exist at $x=0$. But, if I differentiate using first principles instead, I get the derivative as $0$, and so the function IS differentiable at $x=0$. So, is the function differentiable or not, which result do I take?
calculus
calculus
edited Jan 5 at 13:13
Kenny Wong
18.3k21438
asked Jan 5 at 13:09
karun mathewskarun mathews
245
edited Jan 5 at 13:13
Kenny Wong
18.3k21438
asked Jan 5 at 13:09
karun mathewskarun mathews
245
edited Jan 5 at 13:13
Kenny Wong
18.3k21438
edited Jan 5 at 13:13
Kenny Wong
18.3k21438
edited Jan 5 at 13:13
Kenny Wong
18.3k21438
18.3k21438
asked Jan 5 at 13:09
karun mathewskarun mathews
245
asked Jan 5 at 13:09
karun mathewskarun mathews
245
asked Jan 5 at 13:09
karun mathewskarun mathews
245
245
2
$begingroup$
You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
$endgroup$
– metamorphy
Jan 5 at 13:14
1
$begingroup$
To be clear, "differentiable" means "the derivative exists" (which you showed from first principles). It does not mean "the derivative is itself continuous" (which you showed is false here).
$endgroup$
– Mark S.
Jan 5 at 14:05
$begingroup$
But I don't understand, if by differentiation using first principles, I could find the left hand and right hand limits of the derivative, then why is the function still discontinuous? Is it because we can't find the value of the derivative AT x=0?
$endgroup$
– karun mathews
Jan 5 at 14:43
$begingroup$
Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
$endgroup$
– karun mathews
Jan 5 at 14:48
1
$begingroup$
@karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
$endgroup$
– Matt Samuel
Jan 7 at 14:35
|
show 3 more comments
2
$begingroup$
You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
$endgroup$
– metamorphy
Jan 5 at 13:14
1
$begingroup$
To be clear, "differentiable" means "the derivative exists" (which you showed from first principles). It does not mean "the derivative is itself continuous" (which you showed is false here).
$endgroup$
– Mark S.
Jan 5 at 14:05
$begingroup$
But I don't understand, if by differentiation using first principles, I could find the left hand and right hand limits of the derivative, then why is the function still discontinuous? Is it because we can't find the value of the derivative AT x=0?
$endgroup$
– karun mathews
Jan 5 at 14:43
$begingroup$
Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
$endgroup$
– karun mathews
Jan 5 at 14:48
1
$begingroup$
@karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
$endgroup$
– Matt Samuel
Jan 7 at 14:35
2
2
$begingroup$
You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
$endgroup$
– metamorphy
Jan 5 at 13:14
$begingroup$
You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
$endgroup$
– metamorphy
Jan 5 at 13:14
1
1
$begingroup$
To be clear, "differentiable" means "the derivative exists" (which you showed from first principles). It does not mean "the derivative is itself continuous" (which you showed is false here).
$endgroup$
– Mark S.
Jan 5 at 14:05
$begingroup$
To be clear, "differentiable" means "the derivative exists" (which you showed from first principles). It does not mean "the derivative is itself continuous" (which you showed is false here).
$endgroup$
– Mark S.
Jan 5 at 14:05
$begingroup$
But I don't understand, if by differentiation using first principles, I could find the left hand and right hand limits of the derivative, then why is the function still discontinuous? Is it because we can't find the value of the derivative AT x=0?
$endgroup$
– karun mathews
Jan 5 at 14:43
$begingroup$
But I don't understand, if by differentiation using first principles, I could find the left hand and right hand limits of the derivative, then why is the function still discontinuous? Is it because we can't find the value of the derivative AT x=0?
$endgroup$
– karun mathews
Jan 5 at 14:43
$begingroup$
Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
$endgroup$
– karun mathews
Jan 5 at 14:48
$begingroup$
Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
$endgroup$
– karun mathews
Jan 5 at 14:48
1
1
$begingroup$
@karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
$endgroup$
– Matt Samuel
Jan 7 at 14:35
$begingroup$
@karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
$endgroup$
– Matt Samuel
Jan 7 at 14:35
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Compute $$lim_{hto 0}frac{f(0+h)-f(0)}{h}$$ this is $$lim_{hto 0}frac{h^2|cos(frac{pi}{h})|-0}{h}$$ and since $$frac{h^2|cos(frac{pi}{h})|}{h}=h|cos(frac{pi}{h})|le h$$ this tends to zero for $h$ tends to zero.
answered Jan 5 at 13:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
$endgroup$
$begingroup$
He has already computed this.
$endgroup$
– Thomas Shelby
Jan 5 at 13:14
$begingroup$
Ya, I already did this- that's what I meant by differentiation by first principles, sorry I didn't elaborate. My problem is that when I directly differentiate, I get that the derivative doesn't exist, because of that sin(π/x) term- which will become sin(infinity) at x=0.
$endgroup$
– karun mathews
Jan 5 at 13:55
$begingroup$
@karun There's no such thing as $sin(infty)$. All you know is that $xto 0$, $sin(fracpi{x})$ is bounded in absolute value by $1$. This is good enough to show the limit is $0$.
$endgroup$
– Matt Samuel
Jan 5 at 17:30
add a comment |
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1 Answer
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$begingroup$
Hint: Compute $$lim_{hto 0}frac{f(0+h)-f(0)}{h}$$ this is $$lim_{hto 0}frac{h^2|cos(frac{pi}{h})|-0}{h}$$ and since $$frac{h^2|cos(frac{pi}{h})|}{h}=h|cos(frac{pi}{h})|le h$$ this tends to zero for $h$ tends to zero.
answered Jan 5 at 13:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
$endgroup$
$begingroup$
He has already computed this.
$endgroup$
– Thomas Shelby
Jan 5 at 13:14
$begingroup$
Ya, I already did this- that's what I meant by differentiation by first principles, sorry I didn't elaborate. My problem is that when I directly differentiate, I get that the derivative doesn't exist, because of that sin(π/x) term- which will become sin(infinity) at x=0.
$endgroup$
– karun mathews
Jan 5 at 13:55
$begingroup$
@karun There's no such thing as $sin(infty)$. All you know is that $xto 0$, $sin(fracpi{x})$ is bounded in absolute value by $1$. This is good enough to show the limit is $0$.
$endgroup$
– Matt Samuel
Jan 5 at 17:30
add a comment |
$begingroup$
Hint: Compute $$lim_{hto 0}frac{f(0+h)-f(0)}{h}$$ this is $$lim_{hto 0}frac{h^2|cos(frac{pi}{h})|-0}{h}$$ and since $$frac{h^2|cos(frac{pi}{h})|}{h}=h|cos(frac{pi}{h})|le h$$ this tends to zero for $h$ tends to zero.
answered Jan 5 at 13:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
$endgroup$
$begingroup$
He has already computed this.
$endgroup$
– Thomas Shelby
Jan 5 at 13:14
$begingroup$
Ya, I already did this- that's what I meant by differentiation by first principles, sorry I didn't elaborate. My problem is that when I directly differentiate, I get that the derivative doesn't exist, because of that sin(π/x) term- which will become sin(infinity) at x=0.
$endgroup$
– karun mathews
Jan 5 at 13:55
$begingroup$
@karun There's no such thing as $sin(infty)$. All you know is that $xto 0$, $sin(fracpi{x})$ is bounded in absolute value by $1$. This is good enough to show the limit is $0$.
$endgroup$
– Matt Samuel
Jan 5 at 17:30
add a comment |
$begingroup$
Hint: Compute $$lim_{hto 0}frac{f(0+h)-f(0)}{h}$$ this is $$lim_{hto 0}frac{h^2|cos(frac{pi}{h})|-0}{h}$$ and since $$frac{h^2|cos(frac{pi}{h})|}{h}=h|cos(frac{pi}{h})|le h$$ this tends to zero for $h$ tends to zero.
answered Jan 5 at 13:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
$endgroup$
Hint: Compute $$lim_{hto 0}frac{f(0+h)-f(0)}{h}$$ this is $$lim_{hto 0}frac{h^2|cos(frac{pi}{h})|-0}{h}$$ and since $$frac{h^2|cos(frac{pi}{h})|}{h}=h|cos(frac{pi}{h})|le h$$ this tends to zero for $h$ tends to zero.
answered Jan 5 at 13:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
edited Jan 5 at 13:16
answered Jan 5 at 13:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
answered Jan 5 at 13:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
answered Jan 5 at 13:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
73.5k42865
$begingroup$
He has already computed this.
$endgroup$
– Thomas Shelby
Jan 5 at 13:14
$begingroup$
Ya, I already did this- that's what I meant by differentiation by first principles, sorry I didn't elaborate. My problem is that when I directly differentiate, I get that the derivative doesn't exist, because of that sin(π/x) term- which will become sin(infinity) at x=0.
$endgroup$
– karun mathews
Jan 5 at 13:55
$begingroup$
@karun There's no such thing as $sin(infty)$. All you know is that $xto 0$, $sin(fracpi{x})$ is bounded in absolute value by $1$. This is good enough to show the limit is $0$.
$endgroup$
– Matt Samuel
Jan 5 at 17:30
add a comment |
$begingroup$
He has already computed this.
$endgroup$
– Thomas Shelby
Jan 5 at 13:14
$begingroup$
Ya, I already did this- that's what I meant by differentiation by first principles, sorry I didn't elaborate. My problem is that when I directly differentiate, I get that the derivative doesn't exist, because of that sin(π/x) term- which will become sin(infinity) at x=0.
$endgroup$
– karun mathews
Jan 5 at 13:55
$begingroup$
@karun There's no such thing as $sin(infty)$. All you know is that $xto 0$, $sin(fracpi{x})$ is bounded in absolute value by $1$. This is good enough to show the limit is $0$.
$endgroup$
– Matt Samuel
Jan 5 at 17:30
$begingroup$
He has already computed this.
$endgroup$
– Thomas Shelby
Jan 5 at 13:14
$begingroup$
He has already computed this.
$endgroup$
– Thomas Shelby
Jan 5 at 13:14
$begingroup$
Ya, I already did this- that's what I meant by differentiation by first principles, sorry I didn't elaborate. My problem is that when I directly differentiate, I get that the derivative doesn't exist, because of that sin(π/x) term- which will become sin(infinity) at x=0.
$endgroup$
– karun mathews
Jan 5 at 13:55
$begingroup$
Ya, I already did this- that's what I meant by differentiation by first principles, sorry I didn't elaborate. My problem is that when I directly differentiate, I get that the derivative doesn't exist, because of that sin(π/x) term- which will become sin(infinity) at x=0.
$endgroup$
– karun mathews
Jan 5 at 13:55
$begingroup$
@karun There's no such thing as $sin(infty)$. All you know is that $xto 0$, $sin(fracpi{x})$ is bounded in absolute value by $1$. This is good enough to show the limit is $0$.
$endgroup$
– Matt Samuel
Jan 5 at 17:30
$begingroup$
@karun There's no such thing as $sin(infty)$. All you know is that $xto 0$, $sin(fracpi{x})$ is bounded in absolute value by $1$. This is good enough to show the limit is $0$.
$endgroup$
– Matt Samuel
Jan 5 at 17:30
add a comment |
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$begingroup$
You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
$endgroup$
– metamorphy
Jan 5 at 13:14
1
$begingroup$
To be clear, "differentiable" means "the derivative exists" (which you showed from first principles). It does not mean "the derivative is itself continuous" (which you showed is false here).
$endgroup$
– Mark S.
Jan 5 at 14:05
$begingroup$
But I don't understand, if by differentiation using first principles, I could find the left hand and right hand limits of the derivative, then why is the function still discontinuous? Is it because we can't find the value of the derivative AT x=0?
$endgroup$
– karun mathews
Jan 5 at 14:43
$begingroup$
Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
$endgroup$
– karun mathews
Jan 5 at 14:48
1
$begingroup$
@karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
$endgroup$
– Matt Samuel
Jan 7 at 14:35