Finding derivative of $f(x)$ where $f(xy) = f(x) + f(y)$ - without change of variable
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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
asked Nov 26 at 18:22
amir na
385
add a comment |
up vote
1
down vote
favorite
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
asked Nov 26 at 18:22
amir na
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
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
asked Nov 26 at 18:22
amir na
385
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
asked Nov 26 at 18:22
amir na
385
asked Nov 26 at 18:22
amir na
385
asked Nov 26 at 18:22
amir na
385
asked Nov 26 at 18:22
amir na
385
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
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
$$lim_{h to 0} frac{f(x+h) - f(x)}{h} = frac 1x lim_{h to 0} frac{f(1 + h/x)}{h/x} = frac 1x cdot f'(1).$$
answered Nov 26 at 18:42
Umberto P.
38.3k13063
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
|
show 1 more comment
up vote
1
down vote
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}
answered Nov 26 at 18:38
A.Γ.
21.6k22455
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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
$$lim_{h to 0} frac{f(x+h) - f(x)}{h} = frac 1x lim_{h to 0} frac{f(1 + h/x)}{h/x} = frac 1x cdot f'(1).$$
answered Nov 26 at 18:42
Umberto P.
38.3k13063
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
|
show 1 more comment
up vote
2
down vote
accepted
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
$$lim_{h to 0} frac{f(x+h) - f(x)}{h} = frac 1x lim_{h to 0} frac{f(1 + h/x)}{h/x} = frac 1x cdot f'(1).$$
answered Nov 26 at 18:42
Umberto P.
38.3k13063
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
|
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
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
$$lim_{h to 0} frac{f(x+h) - f(x)}{h} = frac 1x lim_{h to 0} frac{f(1 + h/x)}{h/x} = frac 1x cdot f'(1).$$
answered Nov 26 at 18:42
Umberto P.
38.3k13063
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
$$lim_{h to 0} frac{f(x+h) - f(x)}{h} = frac 1x lim_{h to 0} frac{f(1 + h/x)}{h/x} = frac 1x cdot f'(1).$$
answered Nov 26 at 18:42
Umberto P.
38.3k13063
edited Nov 26 at 18:53
answered Nov 26 at 18:42
Umberto P.
38.3k13063
answered Nov 26 at 18:42
Umberto P.
38.3k13063
answered Nov 26 at 18:42
Umberto P.
38.3k13063
38.3k13063
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
|
show 1 more comment
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
|
show 1 more comment
up vote
1
down vote
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}
answered Nov 26 at 18:38
A.Γ.
21.6k22455
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up vote
1
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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}
answered Nov 26 at 18:38
A.Γ.
21.6k22455
add a comment |
up vote
1
down vote
up vote
1
down vote
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}
answered Nov 26 at 18:38
A.Γ.
21.6k22455
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}
answered Nov 26 at 18:38
A.Γ.
21.6k22455
answered Nov 26 at 18:38
A.Γ.
21.6k22455
answered Nov 26 at 18:38
A.Γ.
21.6k22455
answered Nov 26 at 18:38
A.Γ.
21.6k22455
21.6k22455
add a comment |
add a comment |
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@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