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Derivative of a function at a point
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A friend who is still in highschool gave me the following problem. For $x>-1,$ compute $f'(0)$ if $$f(x)=left(x^2-ln^2(x+1)right)^frac13$$ Here is how I solved it. Since it just got messier if I computed the derivative then tried to plug in$x=0$ I used the definition of derivate: $$lim_xto 0fracleft(x^2-ln^2(x+1)right)^frac13-left(0-0right)^frac13x-0=lim_xto 0fracleft(x^2-ln^2(x+1)right)^frac13x.$$ We have the power series of $ln(1+x)=x-fracx^22+O(x^3)rightarrowln^2(1+x)=x^2-x^3+O(x^4)$ so the initial limit is just $$lim_xto 0fracleft(x^3(1-O(x)right)^frac13x=1$$ I couldn't solve it with any other method and my friend didnt learn series yet. Could you help me with an elementary approach for this problem?
limits derivatives
edited Jul 20 at 8:32
Bernard
110k635103
asked Jul 20 at 8:15
Zacky
2,2251326
add a comment |Â
up vote
4
down vote
favorite
A friend who is still in highschool gave me the following problem. For $x>-1,$ compute $f'(0)$ if $$f(x)=left(x^2-ln^2(x+1)right)^frac13$$ Here is how I solved it. Since it just got messier if I computed the derivative then tried to plug in$x=0$ I used the definition of derivate: $$lim_xto 0fracleft(x^2-ln^2(x+1)right)^frac13-left(0-0right)^frac13x-0=lim_xto 0fracleft(x^2-ln^2(x+1)right)^frac13x.$$ We have the power series of $ln(1+x)=x-fracx^22+O(x^3)rightarrowln^2(1+x)=x^2-x^3+O(x^4)$ so the initial limit is just $$lim_xto 0fracleft(x^3(1-O(x)right)^frac13x=1$$ I couldn't solve it with any other method and my friend didnt learn series yet. Could you help me with an elementary approach for this problem?
limits derivatives
edited Jul 20 at 8:32
Bernard
110k635103
asked Jul 20 at 8:15
Zacky
2,2251326
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
A friend who is still in highschool gave me the following problem. For $x>-1,$ compute $f'(0)$ if $$f(x)=left(x^2-ln^2(x+1)right)^frac13$$ Here is how I solved it. Since it just got messier if I computed the derivative then tried to plug in$x=0$ I used the definition of derivate: $$lim_xto 0fracleft(x^2-ln^2(x+1)right)^frac13-left(0-0right)^frac13x-0=lim_xto 0fracleft(x^2-ln^2(x+1)right)^frac13x.$$ We have the power series of $ln(1+x)=x-fracx^22+O(x^3)rightarrowln^2(1+x)=x^2-x^3+O(x^4)$ so the initial limit is just $$lim_xto 0fracleft(x^3(1-O(x)right)^frac13x=1$$ I couldn't solve it with any other method and my friend didnt learn series yet. Could you help me with an elementary approach for this problem?
limits derivatives
edited Jul 20 at 8:32
Bernard
110k635103
asked Jul 20 at 8:15
Zacky
2,2251326
A friend who is still in highschool gave me the following problem. For $x>-1,$ compute $f'(0)$ if $$f(x)=left(x^2-ln^2(x+1)right)^frac13$$ Here is how I solved it. Since it just got messier if I computed the derivative then tried to plug in$x=0$ I used the definition of derivate: $$lim_xto 0fracleft(x^2-ln^2(x+1)right)^frac13-left(0-0right)^frac13x-0=lim_xto 0fracleft(x^2-ln^2(x+1)right)^frac13x.$$ We have the power series of $ln(1+x)=x-fracx^22+O(x^3)rightarrowln^2(1+x)=x^2-x^3+O(x^4)$ so the initial limit is just $$lim_xto 0fracleft(x^3(1-O(x)right)^frac13x=1$$ I couldn't solve it with any other method and my friend didnt learn series yet. Could you help me with an elementary approach for this problem?
limits derivatives
edited Jul 20 at 8:32
Bernard
110k635103
asked Jul 20 at 8:15
Zacky
2,2251326
edited Jul 20 at 8:32
Bernard
110k635103
edited Jul 20 at 8:32
Bernard
110k635103
edited Jul 20 at 8:32
Bernard
110k635103
110k635103
asked Jul 20 at 8:15
Zacky
2,2251326
asked Jul 20 at 8:15
Zacky
2,2251326
asked Jul 20 at 8:15
Zacky
2,2251326
2,2251326
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
4
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accepted
$$lim_xto 0fracleft(x^2-ln^2(x+1)right)^frac13x=l$$
Then by algebraic limit theorem
$$lim_xto 0fracleft(x^2-ln^2(x+1)right)x^3=l^3$$
$$lim_xto 0fracleft(x-ln(x+1)right)x^2lim_xto 0fracleft(x+ln(x+1)right)x=l^3$$
Now using L'Hopital's rule on both parts
$$lim_xto 0fracleft(1-frac1(x+1)right)2xcdotlim_xto 0fracleft(1+frac1x+1right)1=l^3 \ lim_xto0fracx2x(x+1)cdotlim_xto0fracx+2x+1=l^3 \ l^3=1 \l=1$$
answered Jul 20 at 8:42
Piyush Divyanakar
3,258122
4
How do you know that $l$ exists ?
– S.H.W
Jul 20 at 8:53
2
TBH I don't. I was just hoping that it does. This isn't a good method for a proof. The method in the question is a lot better for proof. But this is a good method for someone in highschool who doesn't understand series expansion and stuff like that.
– Piyush Divyanakar
Jul 20 at 8:56
Shouldn't the reasoning be inverted? I.e.: After we calculated $l^3$, we can then conclude that indeed $$lim_xto 0 bigg(fracleft(x^2-ln^2(x+1)right)xbigg)^frac13 = bigg( lim_xto 0fracleft(x^2-ln^2(x+1)right)xbigg)^frac13$$, as $x^frac 1 3$ is continuous.
– Sudix
Jul 20 at 9:25
@PiyushDivyanakar You can rearrange your answer without assuming that the limit exists (see my answer below). I'am ready to remove my answer if you do that.
– Robert Z
Jul 20 at 9:37
@RobertZ I understand. Thank you. :)
– Piyush Divyanakar
Jul 20 at 9:41
add a comment |Â
up vote
5
down vote
This is a rearranged version of Piyush Divyanakar's answer without the assumption that the given limit exists.
Note that by L'Hopital,
$$lim_xto 0fracx+ln(x+1)x=lim_xto 0frac1+1/(x+1)1=2$$
and
$$lim_xto 0fracx-ln(x+1)x^2=lim_xto 0frac1-1/(x+1)2x=lim_xto 0frac1/(x+1)^22=frac12.$$
Hence, as $xto 0$,
$$fracleft(x^2-ln^2(x+1)right)^frac13x=left(fracx+ln(x+1)xcdot fracx-ln(x+1)x^2right)^1/3to left(2cdot frac12right)^1/3=1.$$
Hence
$$f'(0)=lim_xto 0fracf(x)-f(0)x-0=fracleft(x^2-ln^2(x+1)right)^frac13x=1.$$
answered Jul 20 at 9:29
Robert Z
84k954122
If you calculate $f'(x)$ then put $x= 0$ , it is undefined . Why ?
– S.H.W
Jul 20 at 9:36
The definition of derivative at $0$ is $lim_xto 0fracf(x)-f(0)x-0$. Your $f'$ is defined for $x>0$.
– Robert Z
Jul 20 at 9:41
@S.H.W Because $f$ is only one-sided differentiable. When $xleq 0$ the part where $(a/x^3)^1/3=a^1/3/x$ is used is invalid.
– Oppenede
Jul 20 at 9:41
@RobertZ I'm confused , the derivative function shows all of the points where the function is differentiable but here it doesn't work for zero .
– S.H.W
Jul 20 at 9:48
@S.H.W Note that $f$ is a composition and one of the composed functions is $xto x^1/3$ which is not differentiable at $0$. So for $x=0$, here we can't use the chain rule for the derivative.
– Robert Z
Jul 20 at 9:52
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
$$lim_xto 0fracleft(x^2-ln^2(x+1)right)^frac13x=l$$
Then by algebraic limit theorem
$$lim_xto 0fracleft(x^2-ln^2(x+1)right)x^3=l^3$$
$$lim_xto 0fracleft(x-ln(x+1)right)x^2lim_xto 0fracleft(x+ln(x+1)right)x=l^3$$
Now using L'Hopital's rule on both parts
$$lim_xto 0fracleft(1-frac1(x+1)right)2xcdotlim_xto 0fracleft(1+frac1x+1right)1=l^3 \ lim_xto0fracx2x(x+1)cdotlim_xto0fracx+2x+1=l^3 \ l^3=1 \l=1$$
answered Jul 20 at 8:42
Piyush Divyanakar
3,258122
4
How do you know that $l$ exists ?
– S.H.W
Jul 20 at 8:53
2
TBH I don't. I was just hoping that it does. This isn't a good method for a proof. The method in the question is a lot better for proof. But this is a good method for someone in highschool who doesn't understand series expansion and stuff like that.
– Piyush Divyanakar
Jul 20 at 8:56
Shouldn't the reasoning be inverted? I.e.: After we calculated $l^3$, we can then conclude that indeed $$lim_xto 0 bigg(fracleft(x^2-ln^2(x+1)right)xbigg)^frac13 = bigg( lim_xto 0fracleft(x^2-ln^2(x+1)right)xbigg)^frac13$$, as $x^frac 1 3$ is continuous.
– Sudix
Jul 20 at 9:25
@PiyushDivyanakar You can rearrange your answer without assuming that the limit exists (see my answer below). I'am ready to remove my answer if you do that.
– Robert Z
Jul 20 at 9:37
@RobertZ I understand. Thank you. :)
– Piyush Divyanakar
Jul 20 at 9:41
add a comment |Â
up vote
4
down vote
accepted
$$lim_xto 0fracleft(x^2-ln^2(x+1)right)^frac13x=l$$
Then by algebraic limit theorem
$$lim_xto 0fracleft(x^2-ln^2(x+1)right)x^3=l^3$$
$$lim_xto 0fracleft(x-ln(x+1)right)x^2lim_xto 0fracleft(x+ln(x+1)right)x=l^3$$
Now using L'Hopital's rule on both parts
$$lim_xto 0fracleft(1-frac1(x+1)right)2xcdotlim_xto 0fracleft(1+frac1x+1right)1=l^3 \ lim_xto0fracx2x(x+1)cdotlim_xto0fracx+2x+1=l^3 \ l^3=1 \l=1$$
answered Jul 20 at 8:42
Piyush Divyanakar
3,258122
4
How do you know that $l$ exists ?
– S.H.W
Jul 20 at 8:53
2
TBH I don't. I was just hoping that it does. This isn't a good method for a proof. The method in the question is a lot better for proof. But this is a good method for someone in highschool who doesn't understand series expansion and stuff like that.
– Piyush Divyanakar
Jul 20 at 8:56
Shouldn't the reasoning be inverted? I.e.: After we calculated $l^3$, we can then conclude that indeed $$lim_xto 0 bigg(fracleft(x^2-ln^2(x+1)right)xbigg)^frac13 = bigg( lim_xto 0fracleft(x^2-ln^2(x+1)right)xbigg)^frac13$$, as $x^frac 1 3$ is continuous.
– Sudix
Jul 20 at 9:25
@PiyushDivyanakar You can rearrange your answer without assuming that the limit exists (see my answer below). I'am ready to remove my answer if you do that.
– Robert Z
Jul 20 at 9:37
@RobertZ I understand. Thank you. :)
– Piyush Divyanakar
Jul 20 at 9:41
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$$lim_xto 0fracleft(x^2-ln^2(x+1)right)^frac13x=l$$
Then by algebraic limit theorem
$$lim_xto 0fracleft(x^2-ln^2(x+1)right)x^3=l^3$$
$$lim_xto 0fracleft(x-ln(x+1)right)x^2lim_xto 0fracleft(x+ln(x+1)right)x=l^3$$
Now using L'Hopital's rule on both parts
$$lim_xto 0fracleft(1-frac1(x+1)right)2xcdotlim_xto 0fracleft(1+frac1x+1right)1=l^3 \ lim_xto0fracx2x(x+1)cdotlim_xto0fracx+2x+1=l^3 \ l^3=1 \l=1$$
answered Jul 20 at 8:42
Piyush Divyanakar
3,258122
$$lim_xto 0fracleft(x^2-ln^2(x+1)right)^frac13x=l$$
Then by algebraic limit theorem
$$lim_xto 0fracleft(x^2-ln^2(x+1)right)x^3=l^3$$
$$lim_xto 0fracleft(x-ln(x+1)right)x^2lim_xto 0fracleft(x+ln(x+1)right)x=l^3$$
Now using L'Hopital's rule on both parts
$$lim_xto 0fracleft(1-frac1(x+1)right)2xcdotlim_xto 0fracleft(1+frac1x+1right)1=l^3 \ lim_xto0fracx2x(x+1)cdotlim_xto0fracx+2x+1=l^3 \ l^3=1 \l=1$$
answered Jul 20 at 8:42
Piyush Divyanakar
3,258122
answered Jul 20 at 8:42
Piyush Divyanakar
3,258122
answered Jul 20 at 8:42
Piyush Divyanakar
3,258122
answered Jul 20 at 8:42
Piyush Divyanakar
3,258122
3,258122
4
How do you know that $l$ exists ?
– S.H.W
Jul 20 at 8:53
2
TBH I don't. I was just hoping that it does. This isn't a good method for a proof. The method in the question is a lot better for proof. But this is a good method for someone in highschool who doesn't understand series expansion and stuff like that.
– Piyush Divyanakar
Jul 20 at 8:56
Shouldn't the reasoning be inverted? I.e.: After we calculated $l^3$, we can then conclude that indeed $$lim_xto 0 bigg(fracleft(x^2-ln^2(x+1)right)xbigg)^frac13 = bigg( lim_xto 0fracleft(x^2-ln^2(x+1)right)xbigg)^frac13$$, as $x^frac 1 3$ is continuous.
– Sudix
Jul 20 at 9:25
@PiyushDivyanakar You can rearrange your answer without assuming that the limit exists (see my answer below). I'am ready to remove my answer if you do that.
– Robert Z
Jul 20 at 9:37
@RobertZ I understand. Thank you. :)
– Piyush Divyanakar
Jul 20 at 9:41
add a comment |Â
4
How do you know that $l$ exists ?
– S.H.W
Jul 20 at 8:53
2
TBH I don't. I was just hoping that it does. This isn't a good method for a proof. The method in the question is a lot better for proof. But this is a good method for someone in highschool who doesn't understand series expansion and stuff like that.
– Piyush Divyanakar
Jul 20 at 8:56
Shouldn't the reasoning be inverted? I.e.: After we calculated $l^3$, we can then conclude that indeed $$lim_xto 0 bigg(fracleft(x^2-ln^2(x+1)right)xbigg)^frac13 = bigg( lim_xto 0fracleft(x^2-ln^2(x+1)right)xbigg)^frac13$$, as $x^frac 1 3$ is continuous.
– Sudix
Jul 20 at 9:25
@PiyushDivyanakar You can rearrange your answer without assuming that the limit exists (see my answer below). I'am ready to remove my answer if you do that.
– Robert Z
Jul 20 at 9:37
@RobertZ I understand. Thank you. :)
– Piyush Divyanakar
Jul 20 at 9:41
4
4
How do you know that $l$ exists ?
– S.H.W
Jul 20 at 8:53
How do you know that $l$ exists ?
– S.H.W
Jul 20 at 8:53
2
2
TBH I don't. I was just hoping that it does. This isn't a good method for a proof. The method in the question is a lot better for proof. But this is a good method for someone in highschool who doesn't understand series expansion and stuff like that.
– Piyush Divyanakar
Jul 20 at 8:56
TBH I don't. I was just hoping that it does. This isn't a good method for a proof. The method in the question is a lot better for proof. But this is a good method for someone in highschool who doesn't understand series expansion and stuff like that.
– Piyush Divyanakar
Jul 20 at 8:56
Shouldn't the reasoning be inverted? I.e.: After we calculated $l^3$, we can then conclude that indeed $$lim_xto 0 bigg(fracleft(x^2-ln^2(x+1)right)xbigg)^frac13 = bigg( lim_xto 0fracleft(x^2-ln^2(x+1)right)xbigg)^frac13$$, as $x^frac 1 3$ is continuous.
– Sudix
Jul 20 at 9:25
Shouldn't the reasoning be inverted? I.e.: After we calculated $l^3$, we can then conclude that indeed $$lim_xto 0 bigg(fracleft(x^2-ln^2(x+1)right)xbigg)^frac13 = bigg( lim_xto 0fracleft(x^2-ln^2(x+1)right)xbigg)^frac13$$, as $x^frac 1 3$ is continuous.
– Sudix
Jul 20 at 9:25
@PiyushDivyanakar You can rearrange your answer without assuming that the limit exists (see my answer below). I'am ready to remove my answer if you do that.
– Robert Z
Jul 20 at 9:37
@PiyushDivyanakar You can rearrange your answer without assuming that the limit exists (see my answer below). I'am ready to remove my answer if you do that.
– Robert Z
Jul 20 at 9:37
@RobertZ I understand. Thank you. :)
– Piyush Divyanakar
Jul 20 at 9:41
@RobertZ I understand. Thank you. :)
– Piyush Divyanakar
Jul 20 at 9:41
add a comment |Â
up vote
5
down vote
This is a rearranged version of Piyush Divyanakar's answer without the assumption that the given limit exists.
Note that by L'Hopital,
$$lim_xto 0fracx+ln(x+1)x=lim_xto 0frac1+1/(x+1)1=2$$
and
$$lim_xto 0fracx-ln(x+1)x^2=lim_xto 0frac1-1/(x+1)2x=lim_xto 0frac1/(x+1)^22=frac12.$$
Hence, as $xto 0$,
$$fracleft(x^2-ln^2(x+1)right)^frac13x=left(fracx+ln(x+1)xcdot fracx-ln(x+1)x^2right)^1/3to left(2cdot frac12right)^1/3=1.$$
Hence
$$f'(0)=lim_xto 0fracf(x)-f(0)x-0=fracleft(x^2-ln^2(x+1)right)^frac13x=1.$$
answered Jul 20 at 9:29
Robert Z
84k954122
If you calculate $f'(x)$ then put $x= 0$ , it is undefined . Why ?
– S.H.W
Jul 20 at 9:36
The definition of derivative at $0$ is $lim_xto 0fracf(x)-f(0)x-0$. Your $f'$ is defined for $x>0$.
– Robert Z
Jul 20 at 9:41
@S.H.W Because $f$ is only one-sided differentiable. When $xleq 0$ the part where $(a/x^3)^1/3=a^1/3/x$ is used is invalid.
– Oppenede
Jul 20 at 9:41
@RobertZ I'm confused , the derivative function shows all of the points where the function is differentiable but here it doesn't work for zero .
– S.H.W
Jul 20 at 9:48
@S.H.W Note that $f$ is a composition and one of the composed functions is $xto x^1/3$ which is not differentiable at $0$. So for $x=0$, here we can't use the chain rule for the derivative.
– Robert Z
Jul 20 at 9:52
 |Â
show 1 more comment
up vote
5
down vote
This is a rearranged version of Piyush Divyanakar's answer without the assumption that the given limit exists.
Note that by L'Hopital,
$$lim_xto 0fracx+ln(x+1)x=lim_xto 0frac1+1/(x+1)1=2$$
and
$$lim_xto 0fracx-ln(x+1)x^2=lim_xto 0frac1-1/(x+1)2x=lim_xto 0frac1/(x+1)^22=frac12.$$
Hence, as $xto 0$,
$$fracleft(x^2-ln^2(x+1)right)^frac13x=left(fracx+ln(x+1)xcdot fracx-ln(x+1)x^2right)^1/3to left(2cdot frac12right)^1/3=1.$$
Hence
$$f'(0)=lim_xto 0fracf(x)-f(0)x-0=fracleft(x^2-ln^2(x+1)right)^frac13x=1.$$
answered Jul 20 at 9:29
Robert Z
84k954122
If you calculate $f'(x)$ then put $x= 0$ , it is undefined . Why ?
– S.H.W
Jul 20 at 9:36
The definition of derivative at $0$ is $lim_xto 0fracf(x)-f(0)x-0$. Your $f'$ is defined for $x>0$.
– Robert Z
Jul 20 at 9:41
@S.H.W Because $f$ is only one-sided differentiable. When $xleq 0$ the part where $(a/x^3)^1/3=a^1/3/x$ is used is invalid.
– Oppenede
Jul 20 at 9:41
@RobertZ I'm confused , the derivative function shows all of the points where the function is differentiable but here it doesn't work for zero .
– S.H.W
Jul 20 at 9:48
@S.H.W Note that $f$ is a composition and one of the composed functions is $xto x^1/3$ which is not differentiable at $0$. So for $x=0$, here we can't use the chain rule for the derivative.
– Robert Z
Jul 20 at 9:52
 |Â
show 1 more comment
up vote
5
down vote
up vote
5
down vote
This is a rearranged version of Piyush Divyanakar's answer without the assumption that the given limit exists.
Note that by L'Hopital,
$$lim_xto 0fracx+ln(x+1)x=lim_xto 0frac1+1/(x+1)1=2$$
and
$$lim_xto 0fracx-ln(x+1)x^2=lim_xto 0frac1-1/(x+1)2x=lim_xto 0frac1/(x+1)^22=frac12.$$
Hence, as $xto 0$,
$$fracleft(x^2-ln^2(x+1)right)^frac13x=left(fracx+ln(x+1)xcdot fracx-ln(x+1)x^2right)^1/3to left(2cdot frac12right)^1/3=1.$$
Hence
$$f'(0)=lim_xto 0fracf(x)-f(0)x-0=fracleft(x^2-ln^2(x+1)right)^frac13x=1.$$
answered Jul 20 at 9:29
Robert Z
84k954122
This is a rearranged version of Piyush Divyanakar's answer without the assumption that the given limit exists.
Note that by L'Hopital,
$$lim_xto 0fracx+ln(x+1)x=lim_xto 0frac1+1/(x+1)1=2$$
and
$$lim_xto 0fracx-ln(x+1)x^2=lim_xto 0frac1-1/(x+1)2x=lim_xto 0frac1/(x+1)^22=frac12.$$
Hence, as $xto 0$,
$$fracleft(x^2-ln^2(x+1)right)^frac13x=left(fracx+ln(x+1)xcdot fracx-ln(x+1)x^2right)^1/3to left(2cdot frac12right)^1/3=1.$$
Hence
$$f'(0)=lim_xto 0fracf(x)-f(0)x-0=fracleft(x^2-ln^2(x+1)right)^frac13x=1.$$
answered Jul 20 at 9:29
Robert Z
84k954122
edited Jul 20 at 10:02
answered Jul 20 at 9:29
Robert Z
84k954122
answered Jul 20 at 9:29
Robert Z
84k954122
answered Jul 20 at 9:29
Robert Z
84k954122
84k954122
If you calculate $f'(x)$ then put $x= 0$ , it is undefined . Why ?
– S.H.W
Jul 20 at 9:36
The definition of derivative at $0$ is $lim_xto 0fracf(x)-f(0)x-0$. Your $f'$ is defined for $x>0$.
– Robert Z
Jul 20 at 9:41
@S.H.W Because $f$ is only one-sided differentiable. When $xleq 0$ the part where $(a/x^3)^1/3=a^1/3/x$ is used is invalid.
– Oppenede
Jul 20 at 9:41
@RobertZ I'm confused , the derivative function shows all of the points where the function is differentiable but here it doesn't work for zero .
– S.H.W
Jul 20 at 9:48
@S.H.W Note that $f$ is a composition and one of the composed functions is $xto x^1/3$ which is not differentiable at $0$. So for $x=0$, here we can't use the chain rule for the derivative.
– Robert Z
Jul 20 at 9:52
 |Â
show 1 more comment
If you calculate $f'(x)$ then put $x= 0$ , it is undefined . Why ?
– S.H.W
Jul 20 at 9:36
The definition of derivative at $0$ is $lim_xto 0fracf(x)-f(0)x-0$. Your $f'$ is defined for $x>0$.
– Robert Z
Jul 20 at 9:41
@S.H.W Because $f$ is only one-sided differentiable. When $xleq 0$ the part where $(a/x^3)^1/3=a^1/3/x$ is used is invalid.
– Oppenede
Jul 20 at 9:41
@RobertZ I'm confused , the derivative function shows all of the points where the function is differentiable but here it doesn't work for zero .
– S.H.W
Jul 20 at 9:48
@S.H.W Note that $f$ is a composition and one of the composed functions is $xto x^1/3$ which is not differentiable at $0$. So for $x=0$, here we can't use the chain rule for the derivative.
– Robert Z
Jul 20 at 9:52
If you calculate $f'(x)$ then put $x= 0$ , it is undefined . Why ?
– S.H.W
Jul 20 at 9:36
If you calculate $f'(x)$ then put $x= 0$ , it is undefined . Why ?
– S.H.W
Jul 20 at 9:36
The definition of derivative at $0$ is $lim_xto 0fracf(x)-f(0)x-0$. Your $f'$ is defined for $x>0$.
– Robert Z
Jul 20 at 9:41
The definition of derivative at $0$ is $lim_xto 0fracf(x)-f(0)x-0$. Your $f'$ is defined for $x>0$.
– Robert Z
Jul 20 at 9:41
@S.H.W Because $f$ is only one-sided differentiable. When $xleq 0$ the part where $(a/x^3)^1/3=a^1/3/x$ is used is invalid.
– Oppenede
Jul 20 at 9:41
@S.H.W Because $f$ is only one-sided differentiable. When $xleq 0$ the part where $(a/x^3)^1/3=a^1/3/x$ is used is invalid.
– Oppenede
Jul 20 at 9:41
@RobertZ I'm confused , the derivative function shows all of the points where the function is differentiable but here it doesn't work for zero .
– S.H.W
Jul 20 at 9:48
@RobertZ I'm confused , the derivative function shows all of the points where the function is differentiable but here it doesn't work for zero .
– S.H.W
Jul 20 at 9:48
@S.H.W Note that $f$ is a composition and one of the composed functions is $xto x^1/3$ which is not differentiable at $0$. So for $x=0$, here we can't use the chain rule for the derivative.
– Robert Z
Jul 20 at 9:52
@S.H.W Note that $f$ is a composition and one of the composed functions is $xto x^1/3$ which is not differentiable at $0$. So for $x=0$, here we can't use the chain rule for the derivative.
– Robert Z
Jul 20 at 9:52
 |Â
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