Mixing
Finding the derivative of $x^2|x|$
$begingroup$
Usually, for a question involving products, my first approach would be to apply the product rule, giving $frac{d}{dx}(x^2|x|)=x^2(frac{d}{dx}|x|)+|x|2x$.
Using methods discussed here (Finding the Derivative of |x| using the Limit Definition), I got $frac{d}{dx}|x|=frac{x}{|x|}$, so subsituting this in gives $frac{d}{dx}(x^2|x|)=frac{x^3}{|x|}+2x|x|$.
However, the derivative of $x^2|x|$ is in fact equal to $3x|x|$. Where have I gone wrong?
I also tried using $frac{df(x)}{dx}=lim_{xto x_0}frac{f(x)-f(x_0)}{x-x_0}$, which gave $frac{x^2|x|-x_0^2|x_0|}{x-x_0}rightarrowfrac{x^2_0|x_0|-x_0^2|x_0|}{x_0-x_0}=0$ as $xrightarrow x_0$ - I need to get $3x|x|$, so this can't be right!
Any help would be much appreciated.
real-analysis limits derivatives
asked Jan 19 at 17:46
HarmanHarman
302
$endgroup$
add a comment |
$begingroup$
Usually, for a question involving products, my first approach would be to apply the product rule, giving $frac{d}{dx}(x^2|x|)=x^2(frac{d}{dx}|x|)+|x|2x$.
Using methods discussed here (Finding the Derivative of |x| using the Limit Definition), I got $frac{d}{dx}|x|=frac{x}{|x|}$, so subsituting this in gives $frac{d}{dx}(x^2|x|)=frac{x^3}{|x|}+2x|x|$.
However, the derivative of $x^2|x|$ is in fact equal to $3x|x|$. Where have I gone wrong?
I also tried using $frac{df(x)}{dx}=lim_{xto x_0}frac{f(x)-f(x_0)}{x-x_0}$, which gave $frac{x^2|x|-x_0^2|x_0|}{x-x_0}rightarrowfrac{x^2_0|x_0|-x_0^2|x_0|}{x_0-x_0}=0$ as $xrightarrow x_0$ - I need to get $3x|x|$, so this can't be right!
Any help would be much appreciated.
real-analysis limits derivatives
asked Jan 19 at 17:46
HarmanHarman
302
$endgroup$
1
$begingroup$
you have to be careful on where you differentiate. Keep in mind that $mid x mid$ is not differentiable at $0$.
$endgroup$
– J.F
Jan 19 at 17:50
1
$begingroup$
None of this is necessary to compute the derivative $f'(x)$ at $xne0$ since, then, one can get rid of the absolute value. More annoyingly, none of this applies to the derivative at $x=0$ since you are dividing liberally by $|x|$... How do you suggest to find $f'(0)$?
$endgroup$
– Did
Jan 19 at 18:00
add a comment |
$begingroup$
Usually, for a question involving products, my first approach would be to apply the product rule, giving $frac{d}{dx}(x^2|x|)=x^2(frac{d}{dx}|x|)+|x|2x$.
Using methods discussed here (Finding the Derivative of |x| using the Limit Definition), I got $frac{d}{dx}|x|=frac{x}{|x|}$, so subsituting this in gives $frac{d}{dx}(x^2|x|)=frac{x^3}{|x|}+2x|x|$.
However, the derivative of $x^2|x|$ is in fact equal to $3x|x|$. Where have I gone wrong?
I also tried using $frac{df(x)}{dx}=lim_{xto x_0}frac{f(x)-f(x_0)}{x-x_0}$, which gave $frac{x^2|x|-x_0^2|x_0|}{x-x_0}rightarrowfrac{x^2_0|x_0|-x_0^2|x_0|}{x_0-x_0}=0$ as $xrightarrow x_0$ - I need to get $3x|x|$, so this can't be right!
Any help would be much appreciated.
real-analysis limits derivatives
asked Jan 19 at 17:46
HarmanHarman
302
$endgroup$
Usually, for a question involving products, my first approach would be to apply the product rule, giving $frac{d}{dx}(x^2|x|)=x^2(frac{d}{dx}|x|)+|x|2x$.
Using methods discussed here (Finding the Derivative of |x| using the Limit Definition), I got $frac{d}{dx}|x|=frac{x}{|x|}$, so subsituting this in gives $frac{d}{dx}(x^2|x|)=frac{x^3}{|x|}+2x|x|$.
However, the derivative of $x^2|x|$ is in fact equal to $3x|x|$. Where have I gone wrong?
I also tried using $frac{df(x)}{dx}=lim_{xto x_0}frac{f(x)-f(x_0)}{x-x_0}$, which gave $frac{x^2|x|-x_0^2|x_0|}{x-x_0}rightarrowfrac{x^2_0|x_0|-x_0^2|x_0|}{x_0-x_0}=0$ as $xrightarrow x_0$ - I need to get $3x|x|$, so this can't be right!
Any help would be much appreciated.
real-analysis limits derivatives
real-analysis limits derivatives
asked Jan 19 at 17:46
HarmanHarman
302
asked Jan 19 at 17:46
HarmanHarman
302
asked Jan 19 at 17:46
HarmanHarman
302
asked Jan 19 at 17:46
HarmanHarman
302
asked Jan 19 at 17:46
HarmanHarman
302
302
1
$begingroup$
you have to be careful on where you differentiate. Keep in mind that $mid x mid$ is not differentiable at $0$.
$endgroup$
– J.F
Jan 19 at 17:50
1
$begingroup$
None of this is necessary to compute the derivative $f'(x)$ at $xne0$ since, then, one can get rid of the absolute value. More annoyingly, none of this applies to the derivative at $x=0$ since you are dividing liberally by $|x|$... How do you suggest to find $f'(0)$?
$endgroup$
– Did
Jan 19 at 18:00
add a comment |
1
$begingroup$
you have to be careful on where you differentiate. Keep in mind that $mid x mid$ is not differentiable at $0$.
$endgroup$
– J.F
Jan 19 at 17:50
1
$begingroup$
None of this is necessary to compute the derivative $f'(x)$ at $xne0$ since, then, one can get rid of the absolute value. More annoyingly, none of this applies to the derivative at $x=0$ since you are dividing liberally by $|x|$... How do you suggest to find $f'(0)$?
$endgroup$
– Did
Jan 19 at 18:00
1
1
$begingroup$
you have to be careful on where you differentiate. Keep in mind that $mid x mid$ is not differentiable at $0$.
$endgroup$
– J.F
Jan 19 at 17:50
$begingroup$
you have to be careful on where you differentiate. Keep in mind that $mid x mid$ is not differentiable at $0$.
$endgroup$
– J.F
Jan 19 at 17:50
1
1
$begingroup$
None of this is necessary to compute the derivative $f'(x)$ at $xne0$ since, then, one can get rid of the absolute value. More annoyingly, none of this applies to the derivative at $x=0$ since you are dividing liberally by $|x|$... How do you suggest to find $f'(0)$?
$endgroup$
– Did
Jan 19 at 18:00
$begingroup$
None of this is necessary to compute the derivative $f'(x)$ at $xne0$ since, then, one can get rid of the absolute value. More annoyingly, none of this applies to the derivative at $x=0$ since you are dividing liberally by $|x|$... How do you suggest to find $f'(0)$?
$endgroup$
– Did
Jan 19 at 18:00
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Rather than trying to do something clever, why not just apply the definitions directly? First, we can reasonably assume:
Definition: The absolute value function is the function defined by the formula
$$|x| = begin{cases} x & text{if $x ge 0$, and} \ -x & text{ if $x < 0$.} end{cases} $$
(There are other definitions of the absolute value function, but they are all equivalent to this one over $mathbb{R}$, and this one is likely the easiest to use in the context of the current question.)
The function of interest then becomes
$$
f(x)
= x^2|x|
= begin{cases} x^2(x) & text{if $x ge 0$, and} \ x^2(-x) & text{if $x < 0$} end{cases}
= begin{cases} x^3 & text{if $x ge 0$, and} \ -x^3 & text{if $x < 0$.} end{cases}
$$
This can be differentiated piecewise using the power rule to get
$$
f'(x)
= begin{cases} 3x^2 & text{if $x > 0$, and} \ -3x^2 & text{if $x < 0$.} end{cases}
$$
Note that we can't simply "take the derivative" at zero, since zero is not in the interior of either of the two intervals on which we are differentiating. However, at zero we can directly apply the definition of the derivative:
$$ f'(0)
= lim_{hto 0} frac{f(x+h) - f(x)}{h},
$$
if this limit exists. Taking the limit from the right, we get
$$
lim_{hto 0^+} frac{(0+h)^2|0+h| - 0^2|0|}{h}
= lim_{hto 0^+} frac{h^2|h|}{h}
= lim_{hto 0^+} h|h|
= 0.
$$
The limit from the left is essentially the same, from which we may conclude that $f'(0) = 0$. Combining these results, we get
$$
f'(x)
= begin{cases} 3x^2 & text{if $x > 0$} \ 0 & text{if $x=0$, and} \ -3x^2 & text{if $x < 0$.} end{cases}
$$
By again applying the definition of the absolute value function, this can be simplified to $3x|x|$, which is the claimed result.
answered Jan 19 at 18:10
Xander HendersonXander Henderson
14.8k103555
$endgroup$
add a comment |
$begingroup$
Note that$$frac{x^2}{lvert xrvert}=frac{lvert xrvert^2}{lvert xrvert}=lvert xrvert$$and that therefore $dfrac{x^3}{lvert xrvert}=xlvert xrvert$.
answered Jan 19 at 17:51
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
add a comment |
$begingroup$
We can write $x^2$ as $|x|^2$ so we have $dfrac {|x|^2x}{|x|}=x|x|$. Therefore the answer is $x|x|+2x|x|=3x|x|.$
answered Jan 19 at 17:52
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6562625
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add a comment |
$begingroup$
I would write $f(x) = x^2|x| = |x^3|$. Then if $x>0$, $f(x) =x^3$ and $f'(x)3x^2$. If $x<0$ then $f(x) = -x^3$ and $f'(x) = -3x^2$. In both cases $f'(x) = 3x|x|.$
answered Jan 19 at 17:57
B. GoddardB. Goddard
19.3k21442
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Rather than trying to do something clever, why not just apply the definitions directly? First, we can reasonably assume:
Definition: The absolute value function is the function defined by the formula
$$|x| = begin{cases} x & text{if $x ge 0$, and} \ -x & text{ if $x < 0$.} end{cases} $$
(There are other definitions of the absolute value function, but they are all equivalent to this one over $mathbb{R}$, and this one is likely the easiest to use in the context of the current question.)
The function of interest then becomes
$$
f(x)
= x^2|x|
= begin{cases} x^2(x) & text{if $x ge 0$, and} \ x^2(-x) & text{if $x < 0$} end{cases}
= begin{cases} x^3 & text{if $x ge 0$, and} \ -x^3 & text{if $x < 0$.} end{cases}
$$
This can be differentiated piecewise using the power rule to get
$$
f'(x)
= begin{cases} 3x^2 & text{if $x > 0$, and} \ -3x^2 & text{if $x < 0$.} end{cases}
$$
Note that we can't simply "take the derivative" at zero, since zero is not in the interior of either of the two intervals on which we are differentiating. However, at zero we can directly apply the definition of the derivative:
$$ f'(0)
= lim_{hto 0} frac{f(x+h) - f(x)}{h},
$$
if this limit exists. Taking the limit from the right, we get
$$
lim_{hto 0^+} frac{(0+h)^2|0+h| - 0^2|0|}{h}
= lim_{hto 0^+} frac{h^2|h|}{h}
= lim_{hto 0^+} h|h|
= 0.
$$
The limit from the left is essentially the same, from which we may conclude that $f'(0) = 0$. Combining these results, we get
$$
f'(x)
= begin{cases} 3x^2 & text{if $x > 0$} \ 0 & text{if $x=0$, and} \ -3x^2 & text{if $x < 0$.} end{cases}
$$
By again applying the definition of the absolute value function, this can be simplified to $3x|x|$, which is the claimed result.
answered Jan 19 at 18:10
Xander HendersonXander Henderson
14.8k103555
$endgroup$
add a comment |
$begingroup$
Rather than trying to do something clever, why not just apply the definitions directly? First, we can reasonably assume:
Definition: The absolute value function is the function defined by the formula
$$|x| = begin{cases} x & text{if $x ge 0$, and} \ -x & text{ if $x < 0$.} end{cases} $$
(There are other definitions of the absolute value function, but they are all equivalent to this one over $mathbb{R}$, and this one is likely the easiest to use in the context of the current question.)
The function of interest then becomes
$$
f(x)
= x^2|x|
= begin{cases} x^2(x) & text{if $x ge 0$, and} \ x^2(-x) & text{if $x < 0$} end{cases}
= begin{cases} x^3 & text{if $x ge 0$, and} \ -x^3 & text{if $x < 0$.} end{cases}
$$
This can be differentiated piecewise using the power rule to get
$$
f'(x)
= begin{cases} 3x^2 & text{if $x > 0$, and} \ -3x^2 & text{if $x < 0$.} end{cases}
$$
Note that we can't simply "take the derivative" at zero, since zero is not in the interior of either of the two intervals on which we are differentiating. However, at zero we can directly apply the definition of the derivative:
$$ f'(0)
= lim_{hto 0} frac{f(x+h) - f(x)}{h},
$$
if this limit exists. Taking the limit from the right, we get
$$
lim_{hto 0^+} frac{(0+h)^2|0+h| - 0^2|0|}{h}
= lim_{hto 0^+} frac{h^2|h|}{h}
= lim_{hto 0^+} h|h|
= 0.
$$
The limit from the left is essentially the same, from which we may conclude that $f'(0) = 0$. Combining these results, we get
$$
f'(x)
= begin{cases} 3x^2 & text{if $x > 0$} \ 0 & text{if $x=0$, and} \ -3x^2 & text{if $x < 0$.} end{cases}
$$
By again applying the definition of the absolute value function, this can be simplified to $3x|x|$, which is the claimed result.
answered Jan 19 at 18:10
Xander HendersonXander Henderson
14.8k103555
$endgroup$
add a comment |
$begingroup$
Rather than trying to do something clever, why not just apply the definitions directly? First, we can reasonably assume:
Definition: The absolute value function is the function defined by the formula
$$|x| = begin{cases} x & text{if $x ge 0$, and} \ -x & text{ if $x < 0$.} end{cases} $$
(There are other definitions of the absolute value function, but they are all equivalent to this one over $mathbb{R}$, and this one is likely the easiest to use in the context of the current question.)
The function of interest then becomes
$$
f(x)
= x^2|x|
= begin{cases} x^2(x) & text{if $x ge 0$, and} \ x^2(-x) & text{if $x < 0$} end{cases}
= begin{cases} x^3 & text{if $x ge 0$, and} \ -x^3 & text{if $x < 0$.} end{cases}
$$
This can be differentiated piecewise using the power rule to get
$$
f'(x)
= begin{cases} 3x^2 & text{if $x > 0$, and} \ -3x^2 & text{if $x < 0$.} end{cases}
$$
Note that we can't simply "take the derivative" at zero, since zero is not in the interior of either of the two intervals on which we are differentiating. However, at zero we can directly apply the definition of the derivative:
$$ f'(0)
= lim_{hto 0} frac{f(x+h) - f(x)}{h},
$$
if this limit exists. Taking the limit from the right, we get
$$
lim_{hto 0^+} frac{(0+h)^2|0+h| - 0^2|0|}{h}
= lim_{hto 0^+} frac{h^2|h|}{h}
= lim_{hto 0^+} h|h|
= 0.
$$
The limit from the left is essentially the same, from which we may conclude that $f'(0) = 0$. Combining these results, we get
$$
f'(x)
= begin{cases} 3x^2 & text{if $x > 0$} \ 0 & text{if $x=0$, and} \ -3x^2 & text{if $x < 0$.} end{cases}
$$
By again applying the definition of the absolute value function, this can be simplified to $3x|x|$, which is the claimed result.
answered Jan 19 at 18:10
Xander HendersonXander Henderson
14.8k103555
$endgroup$
Rather than trying to do something clever, why not just apply the definitions directly? First, we can reasonably assume:
Definition: The absolute value function is the function defined by the formula
$$|x| = begin{cases} x & text{if $x ge 0$, and} \ -x & text{ if $x < 0$.} end{cases} $$
(There are other definitions of the absolute value function, but they are all equivalent to this one over $mathbb{R}$, and this one is likely the easiest to use in the context of the current question.)
The function of interest then becomes
$$
f(x)
= x^2|x|
= begin{cases} x^2(x) & text{if $x ge 0$, and} \ x^2(-x) & text{if $x < 0$} end{cases}
= begin{cases} x^3 & text{if $x ge 0$, and} \ -x^3 & text{if $x < 0$.} end{cases}
$$
This can be differentiated piecewise using the power rule to get
$$
f'(x)
= begin{cases} 3x^2 & text{if $x > 0$, and} \ -3x^2 & text{if $x < 0$.} end{cases}
$$
Note that we can't simply "take the derivative" at zero, since zero is not in the interior of either of the two intervals on which we are differentiating. However, at zero we can directly apply the definition of the derivative:
$$ f'(0)
= lim_{hto 0} frac{f(x+h) - f(x)}{h},
$$
if this limit exists. Taking the limit from the right, we get
$$
lim_{hto 0^+} frac{(0+h)^2|0+h| - 0^2|0|}{h}
= lim_{hto 0^+} frac{h^2|h|}{h}
= lim_{hto 0^+} h|h|
= 0.
$$
The limit from the left is essentially the same, from which we may conclude that $f'(0) = 0$. Combining these results, we get
$$
f'(x)
= begin{cases} 3x^2 & text{if $x > 0$} \ 0 & text{if $x=0$, and} \ -3x^2 & text{if $x < 0$.} end{cases}
$$
By again applying the definition of the absolute value function, this can be simplified to $3x|x|$, which is the claimed result.
answered Jan 19 at 18:10
Xander HendersonXander Henderson
14.8k103555
answered Jan 19 at 18:10
Xander HendersonXander Henderson
14.8k103555
answered Jan 19 at 18:10
Xander HendersonXander Henderson
14.8k103555
answered Jan 19 at 18:10
Xander HendersonXander Henderson
14.8k103555
14.8k103555
add a comment |
add a comment |
$begingroup$
Note that$$frac{x^2}{lvert xrvert}=frac{lvert xrvert^2}{lvert xrvert}=lvert xrvert$$and that therefore $dfrac{x^3}{lvert xrvert}=xlvert xrvert$.
answered Jan 19 at 17:51
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
add a comment |
$begingroup$
Note that$$frac{x^2}{lvert xrvert}=frac{lvert xrvert^2}{lvert xrvert}=lvert xrvert$$and that therefore $dfrac{x^3}{lvert xrvert}=xlvert xrvert$.
answered Jan 19 at 17:51
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
add a comment |
$begingroup$
Note that$$frac{x^2}{lvert xrvert}=frac{lvert xrvert^2}{lvert xrvert}=lvert xrvert$$and that therefore $dfrac{x^3}{lvert xrvert}=xlvert xrvert$.
answered Jan 19 at 17:51
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
Note that$$frac{x^2}{lvert xrvert}=frac{lvert xrvert^2}{lvert xrvert}=lvert xrvert$$and that therefore $dfrac{x^3}{lvert xrvert}=xlvert xrvert$.
answered Jan 19 at 17:51
José Carlos SantosJosé Carlos Santos
164k22132235
answered Jan 19 at 17:51
José Carlos SantosJosé Carlos Santos
164k22132235
answered Jan 19 at 17:51
José Carlos SantosJosé Carlos Santos
164k22132235
answered Jan 19 at 17:51
José Carlos SantosJosé Carlos Santos
164k22132235
164k22132235
add a comment |
add a comment |
$begingroup$
We can write $x^2$ as $|x|^2$ so we have $dfrac {|x|^2x}{|x|}=x|x|$. Therefore the answer is $x|x|+2x|x|=3x|x|.$
answered Jan 19 at 17:52
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6562625
$endgroup$
add a comment |
$begingroup$
We can write $x^2$ as $|x|^2$ so we have $dfrac {|x|^2x}{|x|}=x|x|$. Therefore the answer is $x|x|+2x|x|=3x|x|.$
answered Jan 19 at 17:52
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6562625
$endgroup$
add a comment |
$begingroup$
We can write $x^2$ as $|x|^2$ so we have $dfrac {|x|^2x}{|x|}=x|x|$. Therefore the answer is $x|x|+2x|x|=3x|x|.$
answered Jan 19 at 17:52
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6562625
$endgroup$
We can write $x^2$ as $|x|^2$ so we have $dfrac {|x|^2x}{|x|}=x|x|$. Therefore the answer is $x|x|+2x|x|=3x|x|.$
answered Jan 19 at 17:52
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6562625
answered Jan 19 at 17:52
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6562625
answered Jan 19 at 17:52
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6562625
answered Jan 19 at 17:52
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6562625
1,6562625
add a comment |
add a comment |
$begingroup$
I would write $f(x) = x^2|x| = |x^3|$. Then if $x>0$, $f(x) =x^3$ and $f'(x)3x^2$. If $x<0$ then $f(x) = -x^3$ and $f'(x) = -3x^2$. In both cases $f'(x) = 3x|x|.$
answered Jan 19 at 17:57
B. GoddardB. Goddard
19.3k21442
$endgroup$
add a comment |
$begingroup$
I would write $f(x) = x^2|x| = |x^3|$. Then if $x>0$, $f(x) =x^3$ and $f'(x)3x^2$. If $x<0$ then $f(x) = -x^3$ and $f'(x) = -3x^2$. In both cases $f'(x) = 3x|x|.$
answered Jan 19 at 17:57
B. GoddardB. Goddard
19.3k21442
$endgroup$
add a comment |
$begingroup$
I would write $f(x) = x^2|x| = |x^3|$. Then if $x>0$, $f(x) =x^3$ and $f'(x)3x^2$. If $x<0$ then $f(x) = -x^3$ and $f'(x) = -3x^2$. In both cases $f'(x) = 3x|x|.$
answered Jan 19 at 17:57
B. GoddardB. Goddard
19.3k21442
$endgroup$
I would write $f(x) = x^2|x| = |x^3|$. Then if $x>0$, $f(x) =x^3$ and $f'(x)3x^2$. If $x<0$ then $f(x) = -x^3$ and $f'(x) = -3x^2$. In both cases $f'(x) = 3x|x|.$
answered Jan 19 at 17:57
B. GoddardB. Goddard
19.3k21442
answered Jan 19 at 17:57
B. GoddardB. Goddard
19.3k21442
answered Jan 19 at 17:57
B. GoddardB. Goddard
19.3k21442
answered Jan 19 at 17:57
B. GoddardB. Goddard
19.3k21442
19.3k21442
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1
$begingroup$
you have to be careful on where you differentiate. Keep in mind that $mid x mid$ is not differentiable at $0$.
$endgroup$
– J.F
Jan 19 at 17:50
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$begingroup$
None of this is necessary to compute the derivative $f'(x)$ at $xne0$ since, then, one can get rid of the absolute value. More annoyingly, none of this applies to the derivative at $x=0$ since you are dividing liberally by $|x|$... How do you suggest to find $f'(0)$?
$endgroup$
– Did
Jan 19 at 18:00