Derivatives of function when the function is defined differently for different values of x
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If $$ f(x) := begin{cases} 0 & x = 0 \ x + x^2sin(1/x) & x neq 0 end{cases} $$ then by the definition of derivative I get $f'(0) = 1$, and if I differentiate $x + x^2sin(1/x)$ and then plug in $0$ I also get $f'(0) = 1$, but I don't know why I can plug in $0$ here since isn't $f(x)=x + x^2sin(1/x)$ only when $xneq0$?
And I'm also wondering since $f(x)=0$ when $x=0$ why can't we just differentiate this function and say $f'(0) = 0$.
Another question I have is that if $$ f'(x) := begin{cases} x^3 & xgeq 0 \ -x^3 & x < 0 end{cases} $$ why do we have to consider $x=0$ to be a separate case when trying to find the derivatives (for $x=0$ we have to use the definition of derivative to show $f'(0) = 0$ but for $xneq0$ we can just differentiate it directly)?
real-analysis calculus functions derivatives definition
edited Dec 16 '18 at 7:45
Robert Z
100k1069140
asked Dec 16 '18 at 7:30
bt203bt203
114
$endgroup$
add a comment |
$begingroup$
If $$ f(x) := begin{cases} 0 & x = 0 \ x + x^2sin(1/x) & x neq 0 end{cases} $$ then by the definition of derivative I get $f'(0) = 1$, and if I differentiate $x + x^2sin(1/x)$ and then plug in $0$ I also get $f'(0) = 1$, but I don't know why I can plug in $0$ here since isn't $f(x)=x + x^2sin(1/x)$ only when $xneq0$?
And I'm also wondering since $f(x)=0$ when $x=0$ why can't we just differentiate this function and say $f'(0) = 0$.
Another question I have is that if $$ f'(x) := begin{cases} x^3 & xgeq 0 \ -x^3 & x < 0 end{cases} $$ why do we have to consider $x=0$ to be a separate case when trying to find the derivatives (for $x=0$ we have to use the definition of derivative to show $f'(0) = 0$ but for $xneq0$ we can just differentiate it directly)?
real-analysis calculus functions derivatives definition
edited Dec 16 '18 at 7:45
Robert Z
100k1069140
asked Dec 16 '18 at 7:30
bt203bt203
114
$endgroup$
3
$begingroup$
The derivative $f'(x)=1+2xsin(1/x)-cos(1/x)$. How did you plug in $0$ to get $f'(0)=1$?
$endgroup$
– Shubham Johri
Dec 16 '18 at 7:33
$begingroup$
For the differentiability at $x=0$ you may use in both cases that if $g$ is a bounded function, defined on a neighbourhood of $0$, then $f(x)=x^2g(x)$ is differentiable in $x=0$: in the first case take $g(x)=sin(1/x)$, in the second one $g(x)=|x|$.
$endgroup$
– Michael Hoppe
Dec 16 '18 at 12:12
add a comment |
$begingroup$
If $$ f(x) := begin{cases} 0 & x = 0 \ x + x^2sin(1/x) & x neq 0 end{cases} $$ then by the definition of derivative I get $f'(0) = 1$, and if I differentiate $x + x^2sin(1/x)$ and then plug in $0$ I also get $f'(0) = 1$, but I don't know why I can plug in $0$ here since isn't $f(x)=x + x^2sin(1/x)$ only when $xneq0$?
And I'm also wondering since $f(x)=0$ when $x=0$ why can't we just differentiate this function and say $f'(0) = 0$.
Another question I have is that if $$ f'(x) := begin{cases} x^3 & xgeq 0 \ -x^3 & x < 0 end{cases} $$ why do we have to consider $x=0$ to be a separate case when trying to find the derivatives (for $x=0$ we have to use the definition of derivative to show $f'(0) = 0$ but for $xneq0$ we can just differentiate it directly)?
real-analysis calculus functions derivatives definition
edited Dec 16 '18 at 7:45
Robert Z
100k1069140
asked Dec 16 '18 at 7:30
bt203bt203
114
$endgroup$
If $$ f(x) := begin{cases} 0 & x = 0 \ x + x^2sin(1/x) & x neq 0 end{cases} $$ then by the definition of derivative I get $f'(0) = 1$, and if I differentiate $x + x^2sin(1/x)$ and then plug in $0$ I also get $f'(0) = 1$, but I don't know why I can plug in $0$ here since isn't $f(x)=x + x^2sin(1/x)$ only when $xneq0$?
And I'm also wondering since $f(x)=0$ when $x=0$ why can't we just differentiate this function and say $f'(0) = 0$.
Another question I have is that if $$ f'(x) := begin{cases} x^3 & xgeq 0 \ -x^3 & x < 0 end{cases} $$ why do we have to consider $x=0$ to be a separate case when trying to find the derivatives (for $x=0$ we have to use the definition of derivative to show $f'(0) = 0$ but for $xneq0$ we can just differentiate it directly)?
real-analysis calculus functions derivatives definition
real-analysis calculus functions derivatives definition
edited Dec 16 '18 at 7:45
Robert Z
100k1069140
asked Dec 16 '18 at 7:30
bt203bt203
114
edited Dec 16 '18 at 7:45
Robert Z
100k1069140
asked Dec 16 '18 at 7:30
bt203bt203
114
edited Dec 16 '18 at 7:45
Robert Z
100k1069140
edited Dec 16 '18 at 7:45
Robert Z
100k1069140
edited Dec 16 '18 at 7:45
Robert Z
100k1069140
100k1069140
asked Dec 16 '18 at 7:30
bt203bt203
114
asked Dec 16 '18 at 7:30
bt203bt203
114
asked Dec 16 '18 at 7:30
bt203bt203
114
114
3
$begingroup$
The derivative $f'(x)=1+2xsin(1/x)-cos(1/x)$. How did you plug in $0$ to get $f'(0)=1$?
$endgroup$
– Shubham Johri
Dec 16 '18 at 7:33
$begingroup$
For the differentiability at $x=0$ you may use in both cases that if $g$ is a bounded function, defined on a neighbourhood of $0$, then $f(x)=x^2g(x)$ is differentiable in $x=0$: in the first case take $g(x)=sin(1/x)$, in the second one $g(x)=|x|$.
$endgroup$
– Michael Hoppe
Dec 16 '18 at 12:12
add a comment |
3
$begingroup$
The derivative $f'(x)=1+2xsin(1/x)-cos(1/x)$. How did you plug in $0$ to get $f'(0)=1$?
$endgroup$
– Shubham Johri
Dec 16 '18 at 7:33
$begingroup$
For the differentiability at $x=0$ you may use in both cases that if $g$ is a bounded function, defined on a neighbourhood of $0$, then $f(x)=x^2g(x)$ is differentiable in $x=0$: in the first case take $g(x)=sin(1/x)$, in the second one $g(x)=|x|$.
$endgroup$
– Michael Hoppe
Dec 16 '18 at 12:12
3
3
$begingroup$
The derivative $f'(x)=1+2xsin(1/x)-cos(1/x)$. How did you plug in $0$ to get $f'(0)=1$?
$endgroup$
– Shubham Johri
Dec 16 '18 at 7:33
$begingroup$
The derivative $f'(x)=1+2xsin(1/x)-cos(1/x)$. How did you plug in $0$ to get $f'(0)=1$?
$endgroup$
– Shubham Johri
Dec 16 '18 at 7:33
$begingroup$
For the differentiability at $x=0$ you may use in both cases that if $g$ is a bounded function, defined on a neighbourhood of $0$, then $f(x)=x^2g(x)$ is differentiable in $x=0$: in the first case take $g(x)=sin(1/x)$, in the second one $g(x)=|x|$.
$endgroup$
– Michael Hoppe
Dec 16 '18 at 12:12
$begingroup$
For the differentiability at $x=0$ you may use in both cases that if $g$ is a bounded function, defined on a neighbourhood of $0$, then $f(x)=x^2g(x)$ is differentiable in $x=0$: in the first case take $g(x)=sin(1/x)$, in the second one $g(x)=|x|$.
$endgroup$
– Michael Hoppe
Dec 16 '18 at 12:12
add a comment |
1 Answer
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$begingroup$
First of all, since $f'(x)=1+2xsinleft(frac1xright)-cosleft(frac1xright)$, you simply cannot plug in $0$.
Concerning the second question, suppose that you define$$begin{array}{rccc}gcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}x&text{ if }xneq0\0&text{ if }x=0.end{cases}end{array}$$If the method that you suggested worked, we would have $g'(0)$. But $g$ is the identity function (that is, $g(x)=x$ for each $x$) and therefore $g'(0)=1$).
FInally, my answer to the second question also answers the third one.
answered Dec 16 '18 at 7:38
José Carlos SantosJosé Carlos Santos
166k22132235
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add a comment |
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$begingroup$
First of all, since $f'(x)=1+2xsinleft(frac1xright)-cosleft(frac1xright)$, you simply cannot plug in $0$.
Concerning the second question, suppose that you define$$begin{array}{rccc}gcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}x&text{ if }xneq0\0&text{ if }x=0.end{cases}end{array}$$If the method that you suggested worked, we would have $g'(0)$. But $g$ is the identity function (that is, $g(x)=x$ for each $x$) and therefore $g'(0)=1$).
FInally, my answer to the second question also answers the third one.
answered Dec 16 '18 at 7:38
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
add a comment |
$begingroup$
First of all, since $f'(x)=1+2xsinleft(frac1xright)-cosleft(frac1xright)$, you simply cannot plug in $0$.
Concerning the second question, suppose that you define$$begin{array}{rccc}gcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}x&text{ if }xneq0\0&text{ if }x=0.end{cases}end{array}$$If the method that you suggested worked, we would have $g'(0)$. But $g$ is the identity function (that is, $g(x)=x$ for each $x$) and therefore $g'(0)=1$).
FInally, my answer to the second question also answers the third one.
answered Dec 16 '18 at 7:38
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
add a comment |
$begingroup$
First of all, since $f'(x)=1+2xsinleft(frac1xright)-cosleft(frac1xright)$, you simply cannot plug in $0$.
Concerning the second question, suppose that you define$$begin{array}{rccc}gcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}x&text{ if }xneq0\0&text{ if }x=0.end{cases}end{array}$$If the method that you suggested worked, we would have $g'(0)$. But $g$ is the identity function (that is, $g(x)=x$ for each $x$) and therefore $g'(0)=1$).
FInally, my answer to the second question also answers the third one.
answered Dec 16 '18 at 7:38
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
First of all, since $f'(x)=1+2xsinleft(frac1xright)-cosleft(frac1xright)$, you simply cannot plug in $0$.
Concerning the second question, suppose that you define$$begin{array}{rccc}gcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}x&text{ if }xneq0\0&text{ if }x=0.end{cases}end{array}$$If the method that you suggested worked, we would have $g'(0)$. But $g$ is the identity function (that is, $g(x)=x$ for each $x$) and therefore $g'(0)=1$).
FInally, my answer to the second question also answers the third one.
answered Dec 16 '18 at 7:38
José Carlos SantosJosé Carlos Santos
166k22132235
answered Dec 16 '18 at 7:38
José Carlos SantosJosé Carlos Santos
166k22132235
answered Dec 16 '18 at 7:38
José Carlos SantosJosé Carlos Santos
166k22132235
answered Dec 16 '18 at 7:38
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
add a comment |
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$begingroup$
The derivative $f'(x)=1+2xsin(1/x)-cos(1/x)$. How did you plug in $0$ to get $f'(0)=1$?
$endgroup$
– Shubham Johri
Dec 16 '18 at 7:33
$begingroup$
For the differentiability at $x=0$ you may use in both cases that if $g$ is a bounded function, defined on a neighbourhood of $0$, then $f(x)=x^2g(x)$ is differentiable in $x=0$: in the first case take $g(x)=sin(1/x)$, in the second one $g(x)=|x|$.
$endgroup$
– Michael Hoppe
Dec 16 '18 at 12:12