Derivative Simplification
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Need help proving the following:
$frac{dp}{d ln(t)} = t frac {dp}{dt} $
This is what I have so far:
Applying chain rule.
$frac{dp}{d ln(t)} = frac {dp}{dt}frac {dt}{d ln(t)} $
Simplifying:
$frac {dp}{dt}frac {dt}{d ln(t)} = frac {dp}{dt}frac {1}{frac{d}{dt}} ln(t)$
Need assistance with the rest.
Thanks!
calculus
asked Dec 10 '18 at 6:21
MrCleanMrClean
84
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add a comment |
$begingroup$
Need help proving the following:
$frac{dp}{d ln(t)} = t frac {dp}{dt} $
This is what I have so far:
Applying chain rule.
$frac{dp}{d ln(t)} = frac {dp}{dt}frac {dt}{d ln(t)} $
Simplifying:
$frac {dp}{dt}frac {dt}{d ln(t)} = frac {dp}{dt}frac {1}{frac{d}{dt}} ln(t)$
Need assistance with the rest.
Thanks!
calculus
asked Dec 10 '18 at 6:21
MrCleanMrClean
84
$endgroup$
$begingroup$
$frac {dt}{d ln(t)}=frac 1{frac {d ln(t)}{dt}}$ could help
$endgroup$
– Claude Leibovici
Dec 10 '18 at 8:35
add a comment |
$begingroup$
Need help proving the following:
$frac{dp}{d ln(t)} = t frac {dp}{dt} $
This is what I have so far:
Applying chain rule.
$frac{dp}{d ln(t)} = frac {dp}{dt}frac {dt}{d ln(t)} $
Simplifying:
$frac {dp}{dt}frac {dt}{d ln(t)} = frac {dp}{dt}frac {1}{frac{d}{dt}} ln(t)$
Need assistance with the rest.
Thanks!
calculus
asked Dec 10 '18 at 6:21
MrCleanMrClean
84
$endgroup$
Need help proving the following:
$frac{dp}{d ln(t)} = t frac {dp}{dt} $
This is what I have so far:
Applying chain rule.
$frac{dp}{d ln(t)} = frac {dp}{dt}frac {dt}{d ln(t)} $
Simplifying:
$frac {dp}{dt}frac {dt}{d ln(t)} = frac {dp}{dt}frac {1}{frac{d}{dt}} ln(t)$
Need assistance with the rest.
Thanks!
calculus
calculus
asked Dec 10 '18 at 6:21
MrCleanMrClean
84
asked Dec 10 '18 at 6:21
MrCleanMrClean
84
asked Dec 10 '18 at 6:21
MrCleanMrClean
84
asked Dec 10 '18 at 6:21
MrCleanMrClean
84
asked Dec 10 '18 at 6:21
MrCleanMrClean
84
84
$begingroup$
$frac {dt}{d ln(t)}=frac 1{frac {d ln(t)}{dt}}$ could help
$endgroup$
– Claude Leibovici
Dec 10 '18 at 8:35
add a comment |
$begingroup$
$frac {dt}{d ln(t)}=frac 1{frac {d ln(t)}{dt}}$ could help
$endgroup$
– Claude Leibovici
Dec 10 '18 at 8:35
$begingroup$
$frac {dt}{d ln(t)}=frac 1{frac {d ln(t)}{dt}}$ could help
$endgroup$
– Claude Leibovici
Dec 10 '18 at 8:35
$begingroup$
$frac {dt}{d ln(t)}=frac 1{frac {d ln(t)}{dt}}$ could help
$endgroup$
– Claude Leibovici
Dec 10 '18 at 8:35
add a comment |
1 Answer
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votes
$begingroup$
Leibniz's notations can sometimes be confusing. Writing something like $frac{1}{frac{mathrm{d}}{mathrm{d}t}}$ is terrible.
Let $p(t) = q(ln(t))$. Then
$$frac{mathrm{d}p(t)}{mathrm{d}ln(t)} = frac{mathrm{d} q(ln(t))}{mathrm{d} ln(t)} = q'(ln(t)).$$ Here we treat $ln(t)$ as a single variable.
Also,
$$frac{mathrm{d}p(t)}{mathrm{d}t} = frac{mathrm{d}q(ln(t))}{mathrm{d}t}= q'(ln(t))(ln(t))'=q'(ln(t))frac{1}{t}.$$
answered Dec 10 '18 at 6:35
tonychow0929tonychow0929
28825
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Leibniz's notations can sometimes be confusing. Writing something like $frac{1}{frac{mathrm{d}}{mathrm{d}t}}$ is terrible.
Let $p(t) = q(ln(t))$. Then
$$frac{mathrm{d}p(t)}{mathrm{d}ln(t)} = frac{mathrm{d} q(ln(t))}{mathrm{d} ln(t)} = q'(ln(t)).$$ Here we treat $ln(t)$ as a single variable.
Also,
$$frac{mathrm{d}p(t)}{mathrm{d}t} = frac{mathrm{d}q(ln(t))}{mathrm{d}t}= q'(ln(t))(ln(t))'=q'(ln(t))frac{1}{t}.$$
answered Dec 10 '18 at 6:35
tonychow0929tonychow0929
28825
$endgroup$
add a comment |
$begingroup$
Leibniz's notations can sometimes be confusing. Writing something like $frac{1}{frac{mathrm{d}}{mathrm{d}t}}$ is terrible.
Let $p(t) = q(ln(t))$. Then
$$frac{mathrm{d}p(t)}{mathrm{d}ln(t)} = frac{mathrm{d} q(ln(t))}{mathrm{d} ln(t)} = q'(ln(t)).$$ Here we treat $ln(t)$ as a single variable.
Also,
$$frac{mathrm{d}p(t)}{mathrm{d}t} = frac{mathrm{d}q(ln(t))}{mathrm{d}t}= q'(ln(t))(ln(t))'=q'(ln(t))frac{1}{t}.$$
answered Dec 10 '18 at 6:35
tonychow0929tonychow0929
28825
$endgroup$
add a comment |
$begingroup$
Leibniz's notations can sometimes be confusing. Writing something like $frac{1}{frac{mathrm{d}}{mathrm{d}t}}$ is terrible.
Let $p(t) = q(ln(t))$. Then
$$frac{mathrm{d}p(t)}{mathrm{d}ln(t)} = frac{mathrm{d} q(ln(t))}{mathrm{d} ln(t)} = q'(ln(t)).$$ Here we treat $ln(t)$ as a single variable.
Also,
$$frac{mathrm{d}p(t)}{mathrm{d}t} = frac{mathrm{d}q(ln(t))}{mathrm{d}t}= q'(ln(t))(ln(t))'=q'(ln(t))frac{1}{t}.$$
answered Dec 10 '18 at 6:35
tonychow0929tonychow0929
28825
$endgroup$
Leibniz's notations can sometimes be confusing. Writing something like $frac{1}{frac{mathrm{d}}{mathrm{d}t}}$ is terrible.
Let $p(t) = q(ln(t))$. Then
$$frac{mathrm{d}p(t)}{mathrm{d}ln(t)} = frac{mathrm{d} q(ln(t))}{mathrm{d} ln(t)} = q'(ln(t)).$$ Here we treat $ln(t)$ as a single variable.
Also,
$$frac{mathrm{d}p(t)}{mathrm{d}t} = frac{mathrm{d}q(ln(t))}{mathrm{d}t}= q'(ln(t))(ln(t))'=q'(ln(t))frac{1}{t}.$$
answered Dec 10 '18 at 6:35
tonychow0929tonychow0929
28825
answered Dec 10 '18 at 6:35
tonychow0929tonychow0929
28825
answered Dec 10 '18 at 6:35
tonychow0929tonychow0929
28825
answered Dec 10 '18 at 6:35
tonychow0929tonychow0929
28825
28825
add a comment |
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$begingroup$
$frac {dt}{d ln(t)}=frac 1{frac {d ln(t)}{dt}}$ could help
$endgroup$
– Claude Leibovici
Dec 10 '18 at 8:35