Trouble checking whether a function is differentiable or not












0














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







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  • 2




    You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
    – metamorphy
    2 days ago








  • 1




    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).
    – Mark S.
    2 days ago










  • 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?
    – karun mathews
    2 days ago










  • Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
    – karun mathews
    2 days ago






  • 1




    @karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
    – Matt Samuel
    4 hours ago
















0














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







share|cite|improve this question




















  • 2




    You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
    – metamorphy
    2 days ago








  • 1




    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).
    – Mark S.
    2 days ago










  • 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?
    – karun mathews
    2 days ago










  • Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
    – karun mathews
    2 days ago






  • 1




    @karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
    – Matt Samuel
    4 hours ago














0












0








0







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







share|cite|improve this question















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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Kenny Wong

18.1k21438




18.1k21438










asked 2 days ago









karun mathewskarun mathews

244




244








  • 2




    You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
    – metamorphy
    2 days ago








  • 1




    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).
    – Mark S.
    2 days ago










  • 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?
    – karun mathews
    2 days ago










  • Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
    – karun mathews
    2 days ago






  • 1




    @karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
    – Matt Samuel
    4 hours ago














  • 2




    You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
    – metamorphy
    2 days ago








  • 1




    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).
    – Mark S.
    2 days ago










  • 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?
    – karun mathews
    2 days ago










  • Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
    – karun mathews
    2 days ago






  • 1




    @karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
    – Matt Samuel
    4 hours ago








2




2




You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
– metamorphy
2 days ago






You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
– metamorphy
2 days ago






1




1




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).
– Mark S.
2 days ago




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).
– Mark S.
2 days ago












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?
– karun mathews
2 days ago




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?
– karun mathews
2 days ago












Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
– karun mathews
2 days ago




Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
– karun mathews
2 days ago




1




1




@karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
– Matt Samuel
4 hours ago




@karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
– Matt Samuel
4 hours ago










1 Answer
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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.






share|cite|improve this answer























  • He has already computed this.
    – Thomas Shelby
    2 days ago










  • 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.
    – karun mathews
    2 days ago










  • @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 $0$.
    – Matt Samuel
    2 days ago











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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.






share|cite|improve this answer























  • He has already computed this.
    – Thomas Shelby
    2 days ago










  • 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.
    – karun mathews
    2 days ago










  • @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 $0$.
    – Matt Samuel
    2 days ago
















0














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.






share|cite|improve this answer























  • He has already computed this.
    – Thomas Shelby
    2 days ago










  • 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.
    – karun mathews
    2 days ago










  • @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 $0$.
    – Matt Samuel
    2 days ago














0












0








0






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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









Dr. Sonnhard GraubnerDr. Sonnhard Graubner

73.5k42865




73.5k42865












  • He has already computed this.
    – Thomas Shelby
    2 days ago










  • 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.
    – karun mathews
    2 days ago










  • @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 $0$.
    – Matt Samuel
    2 days ago


















  • He has already computed this.
    – Thomas Shelby
    2 days ago










  • 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.
    – karun mathews
    2 days ago










  • @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 $0$.
    – Matt Samuel
    2 days ago
















He has already computed this.
– Thomas Shelby
2 days ago




He has already computed this.
– Thomas Shelby
2 days ago












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.
– karun mathews
2 days ago




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.
– karun mathews
2 days ago












@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 $0$.
– Matt Samuel
2 days ago




@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 $0$.
– Matt Samuel
2 days ago


















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