Finding derivative of $f(x)$ where $f(xy) = f(x) + f(y)$ - without change of variable












1














Let $f(x)$ be a function $(0,infty) to R$ and for every $x,y$ in the domain we have: $$f(xy) = f(x) + f(y)$$



It is like logarithm but we don't know the exact form of the function. we know it is differentiable at x=1. Now we want to show it is differentiable at its domain and its derivative is $f'(x) = frac{1}{x} f'(1)$.



Solution:



I can find that $f(1) = 0$ and $f(x/y) = f(x) - f(y)$ So we have:



$f'(1) = lim_{hto 0} frac{f(1+h) - f(1)}{h} = lim_{hto 0}frac{f(1+h)}{h} $



so



$f'(x) = lim_{hto 0} frac{f(x+h) - f(x)}{h} = lim_{hto 0} frac{f(frac{x+h}{x})}{h} = lim_{hto 0} frac{f(1 + h/x)}{h}$.



I know I can solve it with a simple change of variables $h to 0 ~~~rightarrow~~~~ h/x to 0$, But I want to know is there any way to solve this without changing variable? Maybe using definition of limit?






limits derivatives logarithms change-of-variable







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  • 1




    @kimchilover it is stated explicitly that $f$ is differentiable at $x=1$.
    – Umberto P.
    Nov 26 at 18:44












  • @kimchilover it is said in the body of the question that it is differentiable at 1. It is not my deduction.
    – amir na
    Nov 26 at 18:49










  • OK, I understand now.
    – kimchi lover
    Nov 26 at 19:05
















1














Let $f(x)$ be a function $(0,infty) to R$ and for every $x,y$ in the domain we have: $$f(xy) = f(x) + f(y)$$



It is like logarithm but we don't know the exact form of the function. we know it is differentiable at x=1. Now we want to show it is differentiable at its domain and its derivative is $f'(x) = frac{1}{x} f'(1)$.



Solution:



I can find that $f(1) = 0$ and $f(x/y) = f(x) - f(y)$ So we have:



$f'(1) = lim_{hto 0} frac{f(1+h) - f(1)}{h} = lim_{hto 0}frac{f(1+h)}{h} $



so



$f'(x) = lim_{hto 0} frac{f(x+h) - f(x)}{h} = lim_{hto 0} frac{f(frac{x+h}{x})}{h} = lim_{hto 0} frac{f(1 + h/x)}{h}$.



I know I can solve it with a simple change of variables $h to 0 ~~~rightarrow~~~~ h/x to 0$, But I want to know is there any way to solve this without changing variable? Maybe using definition of limit?






limits derivatives logarithms change-of-variable







share|cite|improve this question


















  • 1




    @kimchilover it is stated explicitly that $f$ is differentiable at $x=1$.
    – Umberto P.
    Nov 26 at 18:44












  • @kimchilover it is said in the body of the question that it is differentiable at 1. It is not my deduction.
    – amir na
    Nov 26 at 18:49










  • OK, I understand now.
    – kimchi lover
    Nov 26 at 19:05














1












1








1







Let $f(x)$ be a function $(0,infty) to R$ and for every $x,y$ in the domain we have: $$f(xy) = f(x) + f(y)$$



It is like logarithm but we don't know the exact form of the function. we know it is differentiable at x=1. Now we want to show it is differentiable at its domain and its derivative is $f'(x) = frac{1}{x} f'(1)$.



Solution:



I can find that $f(1) = 0$ and $f(x/y) = f(x) - f(y)$ So we have:



$f'(1) = lim_{hto 0} frac{f(1+h) - f(1)}{h} = lim_{hto 0}frac{f(1+h)}{h} $



so



$f'(x) = lim_{hto 0} frac{f(x+h) - f(x)}{h} = lim_{hto 0} frac{f(frac{x+h}{x})}{h} = lim_{hto 0} frac{f(1 + h/x)}{h}$.



I know I can solve it with a simple change of variables $h to 0 ~~~rightarrow~~~~ h/x to 0$, But I want to know is there any way to solve this without changing variable? Maybe using definition of limit?






limits derivatives logarithms change-of-variable







share|cite|improve this question













Let $f(x)$ be a function $(0,infty) to R$ and for every $x,y$ in the domain we have: $$f(xy) = f(x) + f(y)$$



It is like logarithm but we don't know the exact form of the function. we know it is differentiable at x=1. Now we want to show it is differentiable at its domain and its derivative is $f'(x) = frac{1}{x} f'(1)$.



Solution:



I can find that $f(1) = 0$ and $f(x/y) = f(x) - f(y)$ So we have:



$f'(1) = lim_{hto 0} frac{f(1+h) - f(1)}{h} = lim_{hto 0}frac{f(1+h)}{h} $



so



$f'(x) = lim_{hto 0} frac{f(x+h) - f(x)}{h} = lim_{hto 0} frac{f(frac{x+h}{x})}{h} = lim_{hto 0} frac{f(1 + h/x)}{h}$.



I know I can solve it with a simple change of variables $h to 0 ~~~rightarrow~~~~ h/x to 0$, But I want to know is there any way to solve this without changing variable? Maybe using definition of limit?






limits derivatives logarithms change-of-variable




limits derivatives logarithms change-of-variable






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Nov 26 at 18:22









amir na

385




385








  • 1




    @kimchilover it is stated explicitly that $f$ is differentiable at $x=1$.
    – Umberto P.
    Nov 26 at 18:44












  • @kimchilover it is said in the body of the question that it is differentiable at 1. It is not my deduction.
    – amir na
    Nov 26 at 18:49










  • OK, I understand now.
    – kimchi lover
    Nov 26 at 19:05














  • 1




    @kimchilover it is stated explicitly that $f$ is differentiable at $x=1$.
    – Umberto P.
    Nov 26 at 18:44












  • @kimchilover it is said in the body of the question that it is differentiable at 1. It is not my deduction.
    – amir na
    Nov 26 at 18:49










  • OK, I understand now.
    – kimchi lover
    Nov 26 at 19:05








1




1




@kimchilover it is stated explicitly that $f$ is differentiable at $x=1$.
– Umberto P.
Nov 26 at 18:44






@kimchilover it is stated explicitly that $f$ is differentiable at $x=1$.
– Umberto P.
Nov 26 at 18:44














@kimchilover it is said in the body of the question that it is differentiable at 1. It is not my deduction.
– amir na
Nov 26 at 18:49




@kimchilover it is said in the body of the question that it is differentiable at 1. It is not my deduction.
– amir na
Nov 26 at 18:49












OK, I understand now.
– kimchi lover
Nov 26 at 19:05




OK, I understand now.
– kimchi lover
Nov 26 at 19:05










2 Answers
2






active

oldest

votes


















2














You pointed out that $f'(1) = displaystyle lim_{h to 0} frac{f(1+h)}{h}$ exists using the fact that $f$ is differentiable at $x=1$.



A simple substitution shows that $displaystyle lim_{h to 0} frac{f(1+h/x)}{h/x} = f'(1)$ for all $x > 0$.




The above statement follows from a basic $epsilon$ argument. Let $epsilon > 0$ be given. There exists $delta > 0$ with the property that
$$0 < |h| < delta implies left| frac{f(1+h)}{h} - f'(1) right| < epsilon.$$
Thus
$$0 < |h| < x delta implies 0 < left| frac hx right| < delta implies left| frac{f(1+h/x)}{h/x} - f'(1) right| < epsilon$$ so that $$frac{f(1+h/x)}{h/x} to f'(1).$$




Now let $x > 0$ be arbitrary. Then $$f(x+h) - f(x) = f(x(1 + h/x)) - f(x) = f(1 + h/x)$$ so that $$frac{f(x+h) - f(x)}{h} = frac{f(1 + h/x)}{h} = frac 1x frac{f(1 + h/x)}{h/x}$$
and consequently
$$lim_{h to 0} frac{f(x+h) - f(x)}{h} = frac 1x lim_{h to 0} frac{f(1 + h/x)}{h/x} = frac 1x cdot f'(1).$$






share|cite|improve this answer























  • As I said in the question, I know it can be solved like this, But in this solution it is assumed that $hto0$ means $h/x to 0$. but someone told me when you don't know that the function is continuous, you can't simply substitute variables (i.e. h changed to h/x) . I want to know can it be proved just by definition of limit ($epsilon , delta$)?
    – amir na
    Nov 26 at 18:47










  • Please see the edited answer.
    – Umberto P.
    Nov 26 at 18:49










  • @amirna When $hto 0$ and $xne 0$ then $frac{h}{x}to 0$. It is an easy exercise on $epsilon-delta$: $left|frac{h}{x}right|<epsilon$ iff $|h|<epsilon|x|=delta$.
    – A.Γ.
    Nov 26 at 18:51












  • I saw your answer and I know it is completely valid. However, I want to know whether this question can be solved with definition of limit or not. I mean getting $epsilon,delta$ from f'(1) (because we know it exists) and then changing them somehow for using in the $f'(x)$ definition/
    – amir na
    Nov 26 at 18:54










  • @A.Γ. I saw your comment now, it seems fine but I think maybe it must be $|h/x| < delta '$. because actually $epsilon$ is used in $|f(x) - L| < epsilon$ part of the definition and not in the boundary of the variable.
    – amir na
    Nov 26 at 18:57



















1














Set $y=1$ to verify that $f(1)=0$. Now when $hto 0$
begin{align}
frac{f(x+h)-f(x)}{h}&=frac{f((1+h/x)x)-f(x)}{h}=frac{f(1+h/x)+f(x)-f(x)}{h}=\
&=frac{f(1+h/x)}{h/x}cdot frac{1}{x}=frac{f(1+h/x)-f(1)}{h/x}cdotfrac{1}{x}to ?
end{align}






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    2 Answers
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    2 Answers
    2






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    active

    oldest

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    2














    You pointed out that $f'(1) = displaystyle lim_{h to 0} frac{f(1+h)}{h}$ exists using the fact that $f$ is differentiable at $x=1$.



    A simple substitution shows that $displaystyle lim_{h to 0} frac{f(1+h/x)}{h/x} = f'(1)$ for all $x > 0$.




    The above statement follows from a basic $epsilon$ argument. Let $epsilon > 0$ be given. There exists $delta > 0$ with the property that
    $$0 < |h| < delta implies left| frac{f(1+h)}{h} - f'(1) right| < epsilon.$$
    Thus
    $$0 < |h| < x delta implies 0 < left| frac hx right| < delta implies left| frac{f(1+h/x)}{h/x} - f'(1) right| < epsilon$$ so that $$frac{f(1+h/x)}{h/x} to f'(1).$$




    Now let $x > 0$ be arbitrary. Then $$f(x+h) - f(x) = f(x(1 + h/x)) - f(x) = f(1 + h/x)$$ so that $$frac{f(x+h) - f(x)}{h} = frac{f(1 + h/x)}{h} = frac 1x frac{f(1 + h/x)}{h/x}$$
    and consequently
    $$lim_{h to 0} frac{f(x+h) - f(x)}{h} = frac 1x lim_{h to 0} frac{f(1 + h/x)}{h/x} = frac 1x cdot f'(1).$$






    share|cite|improve this answer























    • As I said in the question, I know it can be solved like this, But in this solution it is assumed that $hto0$ means $h/x to 0$. but someone told me when you don't know that the function is continuous, you can't simply substitute variables (i.e. h changed to h/x) . I want to know can it be proved just by definition of limit ($epsilon , delta$)?
      – amir na
      Nov 26 at 18:47










    • Please see the edited answer.
      – Umberto P.
      Nov 26 at 18:49










    • @amirna When $hto 0$ and $xne 0$ then $frac{h}{x}to 0$. It is an easy exercise on $epsilon-delta$: $left|frac{h}{x}right|<epsilon$ iff $|h|<epsilon|x|=delta$.
      – A.Γ.
      Nov 26 at 18:51












    • I saw your answer and I know it is completely valid. However, I want to know whether this question can be solved with definition of limit or not. I mean getting $epsilon,delta$ from f'(1) (because we know it exists) and then changing them somehow for using in the $f'(x)$ definition/
      – amir na
      Nov 26 at 18:54










    • @A.Γ. I saw your comment now, it seems fine but I think maybe it must be $|h/x| < delta '$. because actually $epsilon$ is used in $|f(x) - L| < epsilon$ part of the definition and not in the boundary of the variable.
      – amir na
      Nov 26 at 18:57
















    2














    You pointed out that $f'(1) = displaystyle lim_{h to 0} frac{f(1+h)}{h}$ exists using the fact that $f$ is differentiable at $x=1$.



    A simple substitution shows that $displaystyle lim_{h to 0} frac{f(1+h/x)}{h/x} = f'(1)$ for all $x > 0$.




    The above statement follows from a basic $epsilon$ argument. Let $epsilon > 0$ be given. There exists $delta > 0$ with the property that
    $$0 < |h| < delta implies left| frac{f(1+h)}{h} - f'(1) right| < epsilon.$$
    Thus
    $$0 < |h| < x delta implies 0 < left| frac hx right| < delta implies left| frac{f(1+h/x)}{h/x} - f'(1) right| < epsilon$$ so that $$frac{f(1+h/x)}{h/x} to f'(1).$$




    Now let $x > 0$ be arbitrary. Then $$f(x+h) - f(x) = f(x(1 + h/x)) - f(x) = f(1 + h/x)$$ so that $$frac{f(x+h) - f(x)}{h} = frac{f(1 + h/x)}{h} = frac 1x frac{f(1 + h/x)}{h/x}$$
    and consequently
    $$lim_{h to 0} frac{f(x+h) - f(x)}{h} = frac 1x lim_{h to 0} frac{f(1 + h/x)}{h/x} = frac 1x cdot f'(1).$$






    share|cite|improve this answer























    • As I said in the question, I know it can be solved like this, But in this solution it is assumed that $hto0$ means $h/x to 0$. but someone told me when you don't know that the function is continuous, you can't simply substitute variables (i.e. h changed to h/x) . I want to know can it be proved just by definition of limit ($epsilon , delta$)?
      – amir na
      Nov 26 at 18:47










    • Please see the edited answer.
      – Umberto P.
      Nov 26 at 18:49










    • @amirna When $hto 0$ and $xne 0$ then $frac{h}{x}to 0$. It is an easy exercise on $epsilon-delta$: $left|frac{h}{x}right|<epsilon$ iff $|h|<epsilon|x|=delta$.
      – A.Γ.
      Nov 26 at 18:51












    • I saw your answer and I know it is completely valid. However, I want to know whether this question can be solved with definition of limit or not. I mean getting $epsilon,delta$ from f'(1) (because we know it exists) and then changing them somehow for using in the $f'(x)$ definition/
      – amir na
      Nov 26 at 18:54










    • @A.Γ. I saw your comment now, it seems fine but I think maybe it must be $|h/x| < delta '$. because actually $epsilon$ is used in $|f(x) - L| < epsilon$ part of the definition and not in the boundary of the variable.
      – amir na
      Nov 26 at 18:57














    2












    2








    2






    You pointed out that $f'(1) = displaystyle lim_{h to 0} frac{f(1+h)}{h}$ exists using the fact that $f$ is differentiable at $x=1$.



    A simple substitution shows that $displaystyle lim_{h to 0} frac{f(1+h/x)}{h/x} = f'(1)$ for all $x > 0$.




    The above statement follows from a basic $epsilon$ argument. Let $epsilon > 0$ be given. There exists $delta > 0$ with the property that
    $$0 < |h| < delta implies left| frac{f(1+h)}{h} - f'(1) right| < epsilon.$$
    Thus
    $$0 < |h| < x delta implies 0 < left| frac hx right| < delta implies left| frac{f(1+h/x)}{h/x} - f'(1) right| < epsilon$$ so that $$frac{f(1+h/x)}{h/x} to f'(1).$$




    Now let $x > 0$ be arbitrary. Then $$f(x+h) - f(x) = f(x(1 + h/x)) - f(x) = f(1 + h/x)$$ so that $$frac{f(x+h) - f(x)}{h} = frac{f(1 + h/x)}{h} = frac 1x frac{f(1 + h/x)}{h/x}$$
    and consequently
    $$lim_{h to 0} frac{f(x+h) - f(x)}{h} = frac 1x lim_{h to 0} frac{f(1 + h/x)}{h/x} = frac 1x cdot f'(1).$$






    share|cite|improve this answer














    You pointed out that $f'(1) = displaystyle lim_{h to 0} frac{f(1+h)}{h}$ exists using the fact that $f$ is differentiable at $x=1$.



    A simple substitution shows that $displaystyle lim_{h to 0} frac{f(1+h/x)}{h/x} = f'(1)$ for all $x > 0$.




    The above statement follows from a basic $epsilon$ argument. Let $epsilon > 0$ be given. There exists $delta > 0$ with the property that
    $$0 < |h| < delta implies left| frac{f(1+h)}{h} - f'(1) right| < epsilon.$$
    Thus
    $$0 < |h| < x delta implies 0 < left| frac hx right| < delta implies left| frac{f(1+h/x)}{h/x} - f'(1) right| < epsilon$$ so that $$frac{f(1+h/x)}{h/x} to f'(1).$$




    Now let $x > 0$ be arbitrary. Then $$f(x+h) - f(x) = f(x(1 + h/x)) - f(x) = f(1 + h/x)$$ so that $$frac{f(x+h) - f(x)}{h} = frac{f(1 + h/x)}{h} = frac 1x frac{f(1 + h/x)}{h/x}$$
    and consequently
    $$lim_{h to 0} frac{f(x+h) - f(x)}{h} = frac 1x lim_{h to 0} frac{f(1 + h/x)}{h/x} = frac 1x cdot f'(1).$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 26 at 18:53

























    answered Nov 26 at 18:42









    Umberto P.

    38.5k13064




    38.5k13064












    • As I said in the question, I know it can be solved like this, But in this solution it is assumed that $hto0$ means $h/x to 0$. but someone told me when you don't know that the function is continuous, you can't simply substitute variables (i.e. h changed to h/x) . I want to know can it be proved just by definition of limit ($epsilon , delta$)?
      – amir na
      Nov 26 at 18:47










    • Please see the edited answer.
      – Umberto P.
      Nov 26 at 18:49










    • @amirna When $hto 0$ and $xne 0$ then $frac{h}{x}to 0$. It is an easy exercise on $epsilon-delta$: $left|frac{h}{x}right|<epsilon$ iff $|h|<epsilon|x|=delta$.
      – A.Γ.
      Nov 26 at 18:51












    • I saw your answer and I know it is completely valid. However, I want to know whether this question can be solved with definition of limit or not. I mean getting $epsilon,delta$ from f'(1) (because we know it exists) and then changing them somehow for using in the $f'(x)$ definition/
      – amir na
      Nov 26 at 18:54










    • @A.Γ. I saw your comment now, it seems fine but I think maybe it must be $|h/x| < delta '$. because actually $epsilon$ is used in $|f(x) - L| < epsilon$ part of the definition and not in the boundary of the variable.
      – amir na
      Nov 26 at 18:57


















    • As I said in the question, I know it can be solved like this, But in this solution it is assumed that $hto0$ means $h/x to 0$. but someone told me when you don't know that the function is continuous, you can't simply substitute variables (i.e. h changed to h/x) . I want to know can it be proved just by definition of limit ($epsilon , delta$)?
      – amir na
      Nov 26 at 18:47










    • Please see the edited answer.
      – Umberto P.
      Nov 26 at 18:49










    • @amirna When $hto 0$ and $xne 0$ then $frac{h}{x}to 0$. It is an easy exercise on $epsilon-delta$: $left|frac{h}{x}right|<epsilon$ iff $|h|<epsilon|x|=delta$.
      – A.Γ.
      Nov 26 at 18:51












    • I saw your answer and I know it is completely valid. However, I want to know whether this question can be solved with definition of limit or not. I mean getting $epsilon,delta$ from f'(1) (because we know it exists) and then changing them somehow for using in the $f'(x)$ definition/
      – amir na
      Nov 26 at 18:54










    • @A.Γ. I saw your comment now, it seems fine but I think maybe it must be $|h/x| < delta '$. because actually $epsilon$ is used in $|f(x) - L| < epsilon$ part of the definition and not in the boundary of the variable.
      – amir na
      Nov 26 at 18:57
















    As I said in the question, I know it can be solved like this, But in this solution it is assumed that $hto0$ means $h/x to 0$. but someone told me when you don't know that the function is continuous, you can't simply substitute variables (i.e. h changed to h/x) . I want to know can it be proved just by definition of limit ($epsilon , delta$)?
    – amir na
    Nov 26 at 18:47




    As I said in the question, I know it can be solved like this, But in this solution it is assumed that $hto0$ means $h/x to 0$. but someone told me when you don't know that the function is continuous, you can't simply substitute variables (i.e. h changed to h/x) . I want to know can it be proved just by definition of limit ($epsilon , delta$)?
    – amir na
    Nov 26 at 18:47












    Please see the edited answer.
    – Umberto P.
    Nov 26 at 18:49




    Please see the edited answer.
    – Umberto P.
    Nov 26 at 18:49












    @amirna When $hto 0$ and $xne 0$ then $frac{h}{x}to 0$. It is an easy exercise on $epsilon-delta$: $left|frac{h}{x}right|<epsilon$ iff $|h|<epsilon|x|=delta$.
    – A.Γ.
    Nov 26 at 18:51






    @amirna When $hto 0$ and $xne 0$ then $frac{h}{x}to 0$. It is an easy exercise on $epsilon-delta$: $left|frac{h}{x}right|<epsilon$ iff $|h|<epsilon|x|=delta$.
    – A.Γ.
    Nov 26 at 18:51














    I saw your answer and I know it is completely valid. However, I want to know whether this question can be solved with definition of limit or not. I mean getting $epsilon,delta$ from f'(1) (because we know it exists) and then changing them somehow for using in the $f'(x)$ definition/
    – amir na
    Nov 26 at 18:54




    I saw your answer and I know it is completely valid. However, I want to know whether this question can be solved with definition of limit or not. I mean getting $epsilon,delta$ from f'(1) (because we know it exists) and then changing them somehow for using in the $f'(x)$ definition/
    – amir na
    Nov 26 at 18:54












    @A.Γ. I saw your comment now, it seems fine but I think maybe it must be $|h/x| < delta '$. because actually $epsilon$ is used in $|f(x) - L| < epsilon$ part of the definition and not in the boundary of the variable.
    – amir na
    Nov 26 at 18:57




    @A.Γ. I saw your comment now, it seems fine but I think maybe it must be $|h/x| < delta '$. because actually $epsilon$ is used in $|f(x) - L| < epsilon$ part of the definition and not in the boundary of the variable.
    – amir na
    Nov 26 at 18:57











    1














    Set $y=1$ to verify that $f(1)=0$. Now when $hto 0$
    begin{align}
    frac{f(x+h)-f(x)}{h}&=frac{f((1+h/x)x)-f(x)}{h}=frac{f(1+h/x)+f(x)-f(x)}{h}=\
    &=frac{f(1+h/x)}{h/x}cdot frac{1}{x}=frac{f(1+h/x)-f(1)}{h/x}cdotfrac{1}{x}to ?
    end{align}






    share|cite|improve this answer


























      1














      Set $y=1$ to verify that $f(1)=0$. Now when $hto 0$
      begin{align}
      frac{f(x+h)-f(x)}{h}&=frac{f((1+h/x)x)-f(x)}{h}=frac{f(1+h/x)+f(x)-f(x)}{h}=\
      &=frac{f(1+h/x)}{h/x}cdot frac{1}{x}=frac{f(1+h/x)-f(1)}{h/x}cdotfrac{1}{x}to ?
      end{align}






      share|cite|improve this answer
























        1












        1








        1






        Set $y=1$ to verify that $f(1)=0$. Now when $hto 0$
        begin{align}
        frac{f(x+h)-f(x)}{h}&=frac{f((1+h/x)x)-f(x)}{h}=frac{f(1+h/x)+f(x)-f(x)}{h}=\
        &=frac{f(1+h/x)}{h/x}cdot frac{1}{x}=frac{f(1+h/x)-f(1)}{h/x}cdotfrac{1}{x}to ?
        end{align}






        share|cite|improve this answer












        Set $y=1$ to verify that $f(1)=0$. Now when $hto 0$
        begin{align}
        frac{f(x+h)-f(x)}{h}&=frac{f((1+h/x)x)-f(x)}{h}=frac{f(1+h/x)+f(x)-f(x)}{h}=\
        &=frac{f(1+h/x)}{h/x}cdot frac{1}{x}=frac{f(1+h/x)-f(1)}{h/x}cdotfrac{1}{x}to ?
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 at 18:38









        A.Γ.

        21.8k22455




        21.8k22455






























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