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On differentiability (product of two functions)

0

$begingroup$

Given that the product $f(x)cdot g(x)$ is differentiable at $x_0$, I need to work out what conditions over $f(x)$ guarantee that $g(x)$ is also differentiable at $x_0$.
Note: both $f$ and $g$ are $mathbbR rightarrow mathbbR$.

My thoughts so far: Well I can easily come up with conditions that guarantee $g(x)$ is differentiable on $x_0$, for example, if $f(x)$ is constant then $(fcdot g)'$ will only exist if $g'$ exists. But that's probably not a necessary condition.

I wrote $(fcdot g)'$ as $ lim_hto 0 left[f(x_0+h)cdotfrac(g(x_0+h)-g(x_0))h +g(x_0)cdotfrac(f(x_0+h)-f(x_0))hright]$ and that gives me a hint that maybe $f(x)$ being differentiable at $x_0$ is sufficient and necessary, but i can't prove it, could anyone please help me out?

calculus derivatives

share|cite|improve this question

asked Mar 25 at 18:16

Leonardo V. SailerLeonardo V. Sailer

305

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Differentiability of $f$ is not enough. For example, take $f(x)=x^2$ and $g(x)=x$ for $x<0$ and $g(x)=2x$ for $xgeq 0$, and work out the derivative of $fcdot g$ at 0 using left/right derivatives.
  $endgroup$
  – Alex R.
  Mar 25 at 18:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I got that wrong, thanks for the counterexample. Can't see what should be the necessary condition then.
  $endgroup$
  – Leonardo V. Sailer
  Mar 25 at 18:40
  

  
  
  
  

add a comment | 

0

$begingroup$

Given that the product $f(x)cdot g(x)$ is differentiable at $x_0$, I need to work out what conditions over $f(x)$ guarantee that $g(x)$ is also differentiable at $x_0$.
Note: both $f$ and $g$ are $mathbbR rightarrow mathbbR$.

My thoughts so far: Well I can easily come up with conditions that guarantee $g(x)$ is differentiable on $x_0$, for example, if $f(x)$ is constant then $(fcdot g)'$ will only exist if $g'$ exists. But that's probably not a necessary condition.

I wrote $(fcdot g)'$ as $ lim_hto 0 left[f(x_0+h)cdotfrac(g(x_0+h)-g(x_0))h +g(x_0)cdotfrac(f(x_0+h)-f(x_0))hright]$ and that gives me a hint that maybe $f(x)$ being differentiable at $x_0$ is sufficient and necessary, but i can't prove it, could anyone please help me out?

calculus derivatives

share|cite|improve this question

asked Mar 25 at 18:16

Leonardo V. SailerLeonardo V. Sailer

305

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Differentiability of $f$ is not enough. For example, take $f(x)=x^2$ and $g(x)=x$ for $x<0$ and $g(x)=2x$ for $xgeq 0$, and work out the derivative of $fcdot g$ at 0 using left/right derivatives.
  $endgroup$
  – Alex R.
  Mar 25 at 18:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I got that wrong, thanks for the counterexample. Can't see what should be the necessary condition then.
  $endgroup$
  – Leonardo V. Sailer
  Mar 25 at 18:40
  

  
  
  
  

add a comment | 

$begingroup$

Given that the product $f(x)cdot g(x)$ is differentiable at $x_0$, I need to work out what conditions over $f(x)$ guarantee that $g(x)$ is also differentiable at $x_0$.
Note: both $f$ and $g$ are $mathbbR rightarrow mathbbR$.

My thoughts so far: Well I can easily come up with conditions that guarantee $g(x)$ is differentiable on $x_0$, for example, if $f(x)$ is constant then $(fcdot g)'$ will only exist if $g'$ exists. But that's probably not a necessary condition.

I wrote $(fcdot g)'$ as $ lim_hto 0 left[f(x_0+h)cdotfrac(g(x_0+h)-g(x_0))h +g(x_0)cdotfrac(f(x_0+h)-f(x_0))hright]$ and that gives me a hint that maybe $f(x)$ being differentiable at $x_0$ is sufficient and necessary, but i can't prove it, could anyone please help me out?

calculus derivatives

share|cite|improve this question

asked Mar 25 at 18:16

Leonardo V. SailerLeonardo V. Sailer

305

$endgroup$

share|cite|improve this question

asked Mar 25 at 18:16

Leonardo V. SailerLeonardo V. Sailer

305

share|cite|improve this question

asked Mar 25 at 18:16

Leonardo V. SailerLeonardo V. Sailer

305

asked Mar 25 at 18:16

Leonardo V. SailerLeonardo V. Sailer

305

asked Mar 25 at 18:16

Leonardo V. SailerLeonardo V. Sailer

305

2 Answers
2

active

oldest

votes

1

$begingroup$

Maybe I'm not seeing the trees for the forest, but to me a 'natural' condition would be that $f$ is differentiable at $x_0$ and $f(x_0) neq 0$. That would make $g$ the quotient of (differentiable at $x_0$) functions $fg$ and $f$, and thus the quotient rule would apply.

The differentiability part for $f$ at $x_0$ is necessary if $g(x_0) neq 0$, with a similar reason.

As has been remarked in the comments, if $f(x_0)=0$, then $g$ may or may not be differentiable (at $x_0$) even if $f$ is.

share|cite|improve this answer

answered Mar 25 at 21:30

IngixIngix

5,232259

$endgroup$

add a comment | 

0

$begingroup$

You must use that
$$lim_hto 0fracg(x_0+h)-g(x_0)h=g'(x_0)$$ and
$$lim_hto 0fracf(x_0+h)-f(x_0)h=f'(x_0)$$
And $$lim_hto 0f(x_0+h)=f(x_0)$$ and $$lim_hto 0g(x_0)=g(x_0)$$

share|cite|improve this answer

edited Mar 25 at 18:44

answered Mar 25 at 18:24

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, doctor, but how can this help on this problem? Since I don't know if the individual limits exist I can't break them apart in two other limits, right?
  $endgroup$
  – Leonardo V. Sailer
  Mar 25 at 18:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given is that $$(fg)$$ is differentiable in $$x_0$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 25 at 18:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  see also here quora.com/How-do-you-prove-the-product-rule-of-differentiation
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 25 at 18:47
  

  
  
  
  

add a comment | 

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1

$begingroup$

Maybe I'm not seeing the trees for the forest, but to me a 'natural' condition would be that $f$ is differentiable at $x_0$ and $f(x_0) neq 0$. That would make $g$ the quotient of (differentiable at $x_0$) functions $fg$ and $f$, and thus the quotient rule would apply.

The differentiability part for $f$ at $x_0$ is necessary if $g(x_0) neq 0$, with a similar reason.

As has been remarked in the comments, if $f(x_0)=0$, then $g$ may or may not be differentiable (at $x_0$) even if $f$ is.

share|cite|improve this answer

answered Mar 25 at 21:30

IngixIngix

5,232259

$endgroup$

add a comment | 

1

$begingroup$

Maybe I'm not seeing the trees for the forest, but to me a 'natural' condition would be that $f$ is differentiable at $x_0$ and $f(x_0) neq 0$. That would make $g$ the quotient of (differentiable at $x_0$) functions $fg$ and $f$, and thus the quotient rule would apply.

The differentiability part for $f$ at $x_0$ is necessary if $g(x_0) neq 0$, with a similar reason.

As has been remarked in the comments, if $f(x_0)=0$, then $g$ may or may not be differentiable (at $x_0$) even if $f$ is.

share|cite|improve this answer

answered Mar 25 at 21:30

IngixIngix

5,232259

$endgroup$

add a comment | 

$begingroup$

Maybe I'm not seeing the trees for the forest, but to me a 'natural' condition would be that $f$ is differentiable at $x_0$ and $f(x_0) neq 0$. That would make $g$ the quotient of (differentiable at $x_0$) functions $fg$ and $f$, and thus the quotient rule would apply.

The differentiability part for $f$ at $x_0$ is necessary if $g(x_0) neq 0$, with a similar reason.

As has been remarked in the comments, if $f(x_0)=0$, then $g$ may or may not be differentiable (at $x_0$) even if $f$ is.

share|cite|improve this answer

answered Mar 25 at 21:30

IngixIngix

5,232259

$endgroup$

share|cite|improve this answer

answered Mar 25 at 21:30

IngixIngix

5,232259

answered Mar 25 at 21:30

IngixIngix

5,232259

answered Mar 25 at 21:30

IngixIngix

5,232259

0

$begingroup$

You must use that
$$lim_hto 0fracg(x_0+h)-g(x_0)h=g'(x_0)$$ and
$$lim_hto 0fracf(x_0+h)-f(x_0)h=f'(x_0)$$
And $$lim_hto 0f(x_0+h)=f(x_0)$$ and $$lim_hto 0g(x_0)=g(x_0)$$

share|cite|improve this answer

edited Mar 25 at 18:44

answered Mar 25 at 18:24

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, doctor, but how can this help on this problem? Since I don't know if the individual limits exist I can't break them apart in two other limits, right?
  $endgroup$
  – Leonardo V. Sailer
  Mar 25 at 18:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given is that $$(fg)$$ is differentiable in $$x_0$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 25 at 18:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  see also here quora.com/How-do-you-prove-the-product-rule-of-differentiation
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 25 at 18:47
  

  
  
  
  

add a comment | 

0

$begingroup$

You must use that
$$lim_hto 0fracg(x_0+h)-g(x_0)h=g'(x_0)$$ and
$$lim_hto 0fracf(x_0+h)-f(x_0)h=f'(x_0)$$
And $$lim_hto 0f(x_0+h)=f(x_0)$$ and $$lim_hto 0g(x_0)=g(x_0)$$

share|cite|improve this answer

edited Mar 25 at 18:44

answered Mar 25 at 18:24

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, doctor, but how can this help on this problem? Since I don't know if the individual limits exist I can't break them apart in two other limits, right?
  $endgroup$
  – Leonardo V. Sailer
  Mar 25 at 18:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given is that $$(fg)$$ is differentiable in $$x_0$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 25 at 18:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  see also here quora.com/How-do-you-prove-the-product-rule-of-differentiation
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 25 at 18:47
  

  
  
  
  

add a comment | 

$begingroup$

You must use that
$$lim_hto 0fracg(x_0+h)-g(x_0)h=g'(x_0)$$ and
$$lim_hto 0fracf(x_0+h)-f(x_0)h=f'(x_0)$$
And $$lim_hto 0f(x_0+h)=f(x_0)$$ and $$lim_hto 0g(x_0)=g(x_0)$$

share|cite|improve this answer

edited Mar 25 at 18:44

answered Mar 25 at 18:24

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

share|cite|improve this answer

edited Mar 25 at 18:44

answered Mar 25 at 18:24

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

answered Mar 25 at 18:24

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

answered Mar 25 at 18:24

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867