Understanding operator under a subtitution
$begingroup$
in my notes, I have the following phrase:
With $x = e^t$ and $x frac{d}{dx} = frac{d}{dt}$
How, how come we get $x frac{d}{dx} = frac{d}{dt} $? I know if I diferentiate with respect to $t$ I obtain
$$ frac{d}{dt} x = x $$
I do not understand how this operators are defined? maybe I am misunderstading the notation?
calculus
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,389319
$endgroup$
add a comment |
$begingroup$
in my notes, I have the following phrase:
With $x = e^t$ and $x frac{d}{dx} = frac{d}{dt}$
How, how come we get $x frac{d}{dx} = frac{d}{dt} $? I know if I diferentiate with respect to $t$ I obtain
$$ frac{d}{dt} x = x $$
I do not understand how this operators are defined? maybe I am misunderstading the notation?
calculus
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,389319
$endgroup$
add a comment |
$begingroup$
in my notes, I have the following phrase:
With $x = e^t$ and $x frac{d}{dx} = frac{d}{dt}$
How, how come we get $x frac{d}{dx} = frac{d}{dt} $? I know if I diferentiate with respect to $t$ I obtain
$$ frac{d}{dt} x = x $$
I do not understand how this operators are defined? maybe I am misunderstading the notation?
calculus
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,389319
$endgroup$
in my notes, I have the following phrase:
With $x = e^t$ and $x frac{d}{dx} = frac{d}{dt}$
How, how come we get $x frac{d}{dx} = frac{d}{dt} $? I know if I diferentiate with respect to $t$ I obtain
$$ frac{d}{dt} x = x $$
I do not understand how this operators are defined? maybe I am misunderstading the notation?
calculus
calculus
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,389319
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,389319
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,389319
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,389319
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,389319
2,389319
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
With
$x = e^t tag 1$
we have
$dfrac{dx}{dt} = e^t = x; tag 2$
thus for any function $f$
$xdfrac{df}{dx} = dfrac{dx}{dt} dfrac{df}{dx} = dfrac{df}{dt} tag 3$
by the chain rule. Thus,
$xdfrac{d}{dx} = dfrac{d}{dt}. tag 4$
answered Jan 9 at 5:33
Robert LewisRobert Lewis
45.1k23065
$endgroup$
add a comment |
$begingroup$
$frac {dy} {dt} =frac {dy} {dx}frac {dx} {dt} =frac {dy} {dx} e^{t}=frac {dy} {dx} x$.
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
With
$x = e^t tag 1$
we have
$dfrac{dx}{dt} = e^t = x; tag 2$
thus for any function $f$
$xdfrac{df}{dx} = dfrac{dx}{dt} dfrac{df}{dx} = dfrac{df}{dt} tag 3$
by the chain rule. Thus,
$xdfrac{d}{dx} = dfrac{d}{dt}. tag 4$
answered Jan 9 at 5:33
Robert LewisRobert Lewis
45.1k23065
$endgroup$
add a comment |
$begingroup$
With
$x = e^t tag 1$
we have
$dfrac{dx}{dt} = e^t = x; tag 2$
thus for any function $f$
$xdfrac{df}{dx} = dfrac{dx}{dt} dfrac{df}{dx} = dfrac{df}{dt} tag 3$
by the chain rule. Thus,
$xdfrac{d}{dx} = dfrac{d}{dt}. tag 4$
answered Jan 9 at 5:33
Robert LewisRobert Lewis
45.1k23065
$endgroup$
add a comment |
$begingroup$
With
$x = e^t tag 1$
we have
$dfrac{dx}{dt} = e^t = x; tag 2$
thus for any function $f$
$xdfrac{df}{dx} = dfrac{dx}{dt} dfrac{df}{dx} = dfrac{df}{dt} tag 3$
by the chain rule. Thus,
$xdfrac{d}{dx} = dfrac{d}{dt}. tag 4$
answered Jan 9 at 5:33
Robert LewisRobert Lewis
45.1k23065
$endgroup$
With
$x = e^t tag 1$
we have
$dfrac{dx}{dt} = e^t = x; tag 2$
thus for any function $f$
$xdfrac{df}{dx} = dfrac{dx}{dt} dfrac{df}{dx} = dfrac{df}{dt} tag 3$
by the chain rule. Thus,
$xdfrac{d}{dx} = dfrac{d}{dt}. tag 4$
answered Jan 9 at 5:33
Robert LewisRobert Lewis
45.1k23065
answered Jan 9 at 5:33
Robert LewisRobert Lewis
45.1k23065
answered Jan 9 at 5:33
Robert LewisRobert Lewis
45.1k23065
answered Jan 9 at 5:33
Robert LewisRobert Lewis
45.1k23065
45.1k23065
add a comment |
add a comment |
$begingroup$
$frac {dy} {dt} =frac {dy} {dx}frac {dx} {dt} =frac {dy} {dx} e^{t}=frac {dy} {dx} x$.
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
add a comment |
$begingroup$
$frac {dy} {dt} =frac {dy} {dx}frac {dx} {dt} =frac {dy} {dx} e^{t}=frac {dy} {dx} x$.
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
add a comment |
$begingroup$
$frac {dy} {dt} =frac {dy} {dx}frac {dx} {dt} =frac {dy} {dx} e^{t}=frac {dy} {dx} x$.
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
$frac {dy} {dt} =frac {dy} {dx}frac {dx} {dt} =frac {dy} {dx} e^{t}=frac {dy} {dx} x$.
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
55.7k42158
add a comment |
add a comment |
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