How we simplify equation with derivation
$begingroup$
Our teacher write :
$(zz')' +2xz' = 0,quad 0 < x < infty tag {1}$
We can write equation $(1)$ like this
$(z^2)'' + 4xz' =0,quad 0 < x < infty tag{2}$
but I can not understand how we find the equation $(2).$
calculus derivatives
edited Nov 30 '18 at 22:16
user376343
3,7883827
asked Nov 30 '18 at 17:46
N111N111
262
$endgroup$
add a comment |
$begingroup$
Our teacher write :
$(zz')' +2xz' = 0,quad 0 < x < infty tag {1}$
We can write equation $(1)$ like this
$(z^2)'' + 4xz' =0,quad 0 < x < infty tag{2}$
but I can not understand how we find the equation $(2).$
calculus derivatives
edited Nov 30 '18 at 22:16
user376343
3,7883827
asked Nov 30 '18 at 17:46
N111N111
262
$endgroup$
$begingroup$
Please write your equations using MathJax. Here's a tutorial.
$endgroup$
– md2perpe
Nov 30 '18 at 17:53
add a comment |
$begingroup$
Our teacher write :
$(zz')' +2xz' = 0,quad 0 < x < infty tag {1}$
We can write equation $(1)$ like this
$(z^2)'' + 4xz' =0,quad 0 < x < infty tag{2}$
but I can not understand how we find the equation $(2).$
calculus derivatives
edited Nov 30 '18 at 22:16
user376343
3,7883827
asked Nov 30 '18 at 17:46
N111N111
262
$endgroup$
Our teacher write :
$(zz')' +2xz' = 0,quad 0 < x < infty tag {1}$
We can write equation $(1)$ like this
$(z^2)'' + 4xz' =0,quad 0 < x < infty tag{2}$
but I can not understand how we find the equation $(2).$
calculus derivatives
calculus derivatives
edited Nov 30 '18 at 22:16
user376343
3,7883827
asked Nov 30 '18 at 17:46
N111N111
262
edited Nov 30 '18 at 22:16
user376343
3,7883827
asked Nov 30 '18 at 17:46
N111N111
262
edited Nov 30 '18 at 22:16
user376343
3,7883827
edited Nov 30 '18 at 22:16
user376343
3,7883827
edited Nov 30 '18 at 22:16
user376343
3,7883827
3,7883827
asked Nov 30 '18 at 17:46
N111N111
262
asked Nov 30 '18 at 17:46
N111N111
262
asked Nov 30 '18 at 17:46
N111N111
262
262
$begingroup$
Please write your equations using MathJax. Here's a tutorial.
$endgroup$
– md2perpe
Nov 30 '18 at 17:53
add a comment |
$begingroup$
Please write your equations using MathJax. Here's a tutorial.
$endgroup$
– md2perpe
Nov 30 '18 at 17:53
$begingroup$
Please write your equations using MathJax. Here's a tutorial.
$endgroup$
– md2perpe
Nov 30 '18 at 17:53
$begingroup$
Please write your equations using MathJax. Here's a tutorial.
$endgroup$
– md2perpe
Nov 30 '18 at 17:53
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The key observation here is that
$$
(z^2)'' = 2(zz')'
$$
which can be derived using either the product rule or the chain rule. Using the chain rule, if we let $f(x) = x^2$, then we're looking to find $(f(z))' = f'(z)z'$, and since $f'(x) = 2x$, this is just $2zz'$. To get the second derivative, just apply the derivative again (but don't simplify): $(f(z))'' = (f'(z)z')' = (2zz')' = 2(zz')'$, where the last equality is true since the derivative is a linear operator.
This means, if we multiply the first equation through by $2$, then we get
$$
2(zz')' + 4xz' = (z^2)' + 4xz'
$$
as expected.
answered Nov 30 '18 at 17:57
user3002473user3002473
3,38752239
$endgroup$
add a comment |
$begingroup$
Application of the Chain Rule :
$$(z^2)’ = 2zz’$$
Thus:
$$(z^2)’ + 4xz = 0$$
becomes
$$2(zz’)’ + 4xz =0$$
Divide through by 2 to get your first equation.
answered Nov 30 '18 at 17:59
ip6ip6
55039
$endgroup$
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
$begingroup$
The key observation here is that
$$
(z^2)'' = 2(zz')'
$$
which can be derived using either the product rule or the chain rule. Using the chain rule, if we let $f(x) = x^2$, then we're looking to find $(f(z))' = f'(z)z'$, and since $f'(x) = 2x$, this is just $2zz'$. To get the second derivative, just apply the derivative again (but don't simplify): $(f(z))'' = (f'(z)z')' = (2zz')' = 2(zz')'$, where the last equality is true since the derivative is a linear operator.
This means, if we multiply the first equation through by $2$, then we get
$$
2(zz')' + 4xz' = (z^2)' + 4xz'
$$
as expected.
answered Nov 30 '18 at 17:57
user3002473user3002473
3,38752239
$endgroup$
add a comment |
$begingroup$
The key observation here is that
$$
(z^2)'' = 2(zz')'
$$
which can be derived using either the product rule or the chain rule. Using the chain rule, if we let $f(x) = x^2$, then we're looking to find $(f(z))' = f'(z)z'$, and since $f'(x) = 2x$, this is just $2zz'$. To get the second derivative, just apply the derivative again (but don't simplify): $(f(z))'' = (f'(z)z')' = (2zz')' = 2(zz')'$, where the last equality is true since the derivative is a linear operator.
This means, if we multiply the first equation through by $2$, then we get
$$
2(zz')' + 4xz' = (z^2)' + 4xz'
$$
as expected.
answered Nov 30 '18 at 17:57
user3002473user3002473
3,38752239
$endgroup$
add a comment |
$begingroup$
The key observation here is that
$$
(z^2)'' = 2(zz')'
$$
which can be derived using either the product rule or the chain rule. Using the chain rule, if we let $f(x) = x^2$, then we're looking to find $(f(z))' = f'(z)z'$, and since $f'(x) = 2x$, this is just $2zz'$. To get the second derivative, just apply the derivative again (but don't simplify): $(f(z))'' = (f'(z)z')' = (2zz')' = 2(zz')'$, where the last equality is true since the derivative is a linear operator.
This means, if we multiply the first equation through by $2$, then we get
$$
2(zz')' + 4xz' = (z^2)' + 4xz'
$$
as expected.
answered Nov 30 '18 at 17:57
user3002473user3002473
3,38752239
$endgroup$
The key observation here is that
$$
(z^2)'' = 2(zz')'
$$
which can be derived using either the product rule or the chain rule. Using the chain rule, if we let $f(x) = x^2$, then we're looking to find $(f(z))' = f'(z)z'$, and since $f'(x) = 2x$, this is just $2zz'$. To get the second derivative, just apply the derivative again (but don't simplify): $(f(z))'' = (f'(z)z')' = (2zz')' = 2(zz')'$, where the last equality is true since the derivative is a linear operator.
This means, if we multiply the first equation through by $2$, then we get
$$
2(zz')' + 4xz' = (z^2)' + 4xz'
$$
as expected.
answered Nov 30 '18 at 17:57
user3002473user3002473
3,38752239
answered Nov 30 '18 at 17:57
user3002473user3002473
3,38752239
answered Nov 30 '18 at 17:57
user3002473user3002473
3,38752239
answered Nov 30 '18 at 17:57
user3002473user3002473
3,38752239
3,38752239
add a comment |
add a comment |
$begingroup$
Application of the Chain Rule :
$$(z^2)’ = 2zz’$$
Thus:
$$(z^2)’ + 4xz = 0$$
becomes
$$2(zz’)’ + 4xz =0$$
Divide through by 2 to get your first equation.
answered Nov 30 '18 at 17:59
ip6ip6
55039
$endgroup$
add a comment |
$begingroup$
Application of the Chain Rule :
$$(z^2)’ = 2zz’$$
Thus:
$$(z^2)’ + 4xz = 0$$
becomes
$$2(zz’)’ + 4xz =0$$
Divide through by 2 to get your first equation.
answered Nov 30 '18 at 17:59
ip6ip6
55039
$endgroup$
add a comment |
$begingroup$
Application of the Chain Rule :
$$(z^2)’ = 2zz’$$
Thus:
$$(z^2)’ + 4xz = 0$$
becomes
$$2(zz’)’ + 4xz =0$$
Divide through by 2 to get your first equation.
answered Nov 30 '18 at 17:59
ip6ip6
55039
$endgroup$
Application of the Chain Rule :
$$(z^2)’ = 2zz’$$
Thus:
$$(z^2)’ + 4xz = 0$$
becomes
$$2(zz’)’ + 4xz =0$$
Divide through by 2 to get your first equation.
answered Nov 30 '18 at 17:59
ip6ip6
55039
answered Nov 30 '18 at 17:59
ip6ip6
55039
answered Nov 30 '18 at 17:59
ip6ip6
55039
answered Nov 30 '18 at 17:59
ip6ip6
55039
55039
add a comment |
add a comment |
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$begingroup$
Please write your equations using MathJax. Here's a tutorial.
$endgroup$
– md2perpe
Nov 30 '18 at 17:53