Mixing
Taking derivative with quotient rule and sums
$begingroup$
I am trying to differentiate this with respect to $eta$ but am having some difficulty.
begin{equation}
g_0(1,eta) = frac{frac{3eta}{(eta_c - eta)} + sumlimits_{k=1}^{4}kA_kleft(frac{eta}{eta_c}right)^k}{4eta}
end{equation}
begin{equation}
g_0'(1,eta) = frac{(4eta)left(frac{3eta_c}{(eta_c - eta)^2}+sumlimits_{k=1}^{4}k^2A_kleft(frac{eta}{eta_c}right)^{k-1}right)-left(frac{3eta}{(eta_c - eta)} + sumlimits_{k=1}^{4}kA_kleft(frac{eta}{eta_c}right)^kright)(4)}{(4eta)^2}
end{equation}
Does this look right? When I continue from here, it just does not seem like anything is cancelling out or anything like that, so I want to make sure what I am doing looks correct.
derivatives
asked Jan 15 at 18:47
Jackson HartJackson Hart
5162626
$endgroup$
add a comment |
$begingroup$
I am trying to differentiate this with respect to $eta$ but am having some difficulty.
begin{equation}
g_0(1,eta) = frac{frac{3eta}{(eta_c - eta)} + sumlimits_{k=1}^{4}kA_kleft(frac{eta}{eta_c}right)^k}{4eta}
end{equation}
begin{equation}
g_0'(1,eta) = frac{(4eta)left(frac{3eta_c}{(eta_c - eta)^2}+sumlimits_{k=1}^{4}k^2A_kleft(frac{eta}{eta_c}right)^{k-1}right)-left(frac{3eta}{(eta_c - eta)} + sumlimits_{k=1}^{4}kA_kleft(frac{eta}{eta_c}right)^kright)(4)}{(4eta)^2}
end{equation}
Does this look right? When I continue from here, it just does not seem like anything is cancelling out or anything like that, so I want to make sure what I am doing looks correct.
derivatives
asked Jan 15 at 18:47
Jackson HartJackson Hart
5162626
$endgroup$
add a comment |
$begingroup$
I am trying to differentiate this with respect to $eta$ but am having some difficulty.
begin{equation}
g_0(1,eta) = frac{frac{3eta}{(eta_c - eta)} + sumlimits_{k=1}^{4}kA_kleft(frac{eta}{eta_c}right)^k}{4eta}
end{equation}
begin{equation}
g_0'(1,eta) = frac{(4eta)left(frac{3eta_c}{(eta_c - eta)^2}+sumlimits_{k=1}^{4}k^2A_kleft(frac{eta}{eta_c}right)^{k-1}right)-left(frac{3eta}{(eta_c - eta)} + sumlimits_{k=1}^{4}kA_kleft(frac{eta}{eta_c}right)^kright)(4)}{(4eta)^2}
end{equation}
Does this look right? When I continue from here, it just does not seem like anything is cancelling out or anything like that, so I want to make sure what I am doing looks correct.
derivatives
asked Jan 15 at 18:47
Jackson HartJackson Hart
5162626
$endgroup$
I am trying to differentiate this with respect to $eta$ but am having some difficulty.
begin{equation}
g_0(1,eta) = frac{frac{3eta}{(eta_c - eta)} + sumlimits_{k=1}^{4}kA_kleft(frac{eta}{eta_c}right)^k}{4eta}
end{equation}
begin{equation}
g_0'(1,eta) = frac{(4eta)left(frac{3eta_c}{(eta_c - eta)^2}+sumlimits_{k=1}^{4}k^2A_kleft(frac{eta}{eta_c}right)^{k-1}right)-left(frac{3eta}{(eta_c - eta)} + sumlimits_{k=1}^{4}kA_kleft(frac{eta}{eta_c}right)^kright)(4)}{(4eta)^2}
end{equation}
Does this look right? When I continue from here, it just does not seem like anything is cancelling out or anything like that, so I want to make sure what I am doing looks correct.
derivatives
derivatives
asked Jan 15 at 18:47
Jackson HartJackson Hart
5162626
asked Jan 15 at 18:47
Jackson HartJackson Hart
5162626
asked Jan 15 at 18:47
Jackson HartJackson Hart
5162626
asked Jan 15 at 18:47
Jackson HartJackson Hart
5162626
asked Jan 15 at 18:47
Jackson HartJackson Hart
5162626
5162626
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
begin{eqnarray}
require{cancel}
frac{{rm d}}{{rm d}eta}g(1, eta) &=& frac{1}{4}frac{{rm d}}{{rm d}eta} left[ frac{3cancel{eta}}{cancel{eta}(eta_c - eta)} + sum_{k=1}^4 k A_k frac{eta^k}{eta eta_c^k}right] \
&=& frac{3}{4} frac{{rm d}}{{rm d}eta} frac{1}{eta_c - eta} + frac{1}{4}sum_{k=1}^4 frac{k A_k}{eta_c^k} frac{{rm d}eta^{k-1}}{{rm d}eta} \
&=& frac{3}{4}frac{1}{(eta_c - eta)^2} + frac{1}{4}sum_{k=1}^4frac{k(k-1) A_k}{eta_c^k}eta^{k - 2} \
&=& frac{3}{4}frac{1}{(eta_c - eta)^2} + frac{1}{4}sum_{k=1}^4frac{k(k-1) A_k}{eta_c^2}left(frac{eta}{eta_c}right)^{k - 2}
end{eqnarray}
answered Jan 15 at 19:13
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Wow, thanks! As a follow-up, assuming that $x=frac{eta}{eta_c}$, how can I get rid of all $eta$'s and just express them as x?
$endgroup$
– Jackson Hart
Jan 15 at 20:16
$begingroup$
The first term becomes $$ frac{3}{4}frac{1}{eta_c^2(1 - eta/eta_c)^2} = frac{3}{4eta_c^2} frac{1}{(1 - x)^2} $$ the last one is the same except replace the factor in parenthesis by $x$
$endgroup$
– caverac
Jan 15 at 20:24
$begingroup$
Did you accidentally lose a $eta$? The last term in first and second lines. In first line, there is an $eta$ in denominator. Where is it on next line? After the $d eta$ shouldnt there also be an $eta$ in denominator? It looks like you dropped it?
$endgroup$
– Jackson Hart
Jan 15 at 20:48
$begingroup$
@JacksonHart Yes I did! Sorry about that. Fixed
$endgroup$
– caverac
Jan 15 at 20:52
add a comment |
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1 Answer
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1 Answer
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$begingroup$
begin{eqnarray}
require{cancel}
frac{{rm d}}{{rm d}eta}g(1, eta) &=& frac{1}{4}frac{{rm d}}{{rm d}eta} left[ frac{3cancel{eta}}{cancel{eta}(eta_c - eta)} + sum_{k=1}^4 k A_k frac{eta^k}{eta eta_c^k}right] \
&=& frac{3}{4} frac{{rm d}}{{rm d}eta} frac{1}{eta_c - eta} + frac{1}{4}sum_{k=1}^4 frac{k A_k}{eta_c^k} frac{{rm d}eta^{k-1}}{{rm d}eta} \
&=& frac{3}{4}frac{1}{(eta_c - eta)^2} + frac{1}{4}sum_{k=1}^4frac{k(k-1) A_k}{eta_c^k}eta^{k - 2} \
&=& frac{3}{4}frac{1}{(eta_c - eta)^2} + frac{1}{4}sum_{k=1}^4frac{k(k-1) A_k}{eta_c^2}left(frac{eta}{eta_c}right)^{k - 2}
end{eqnarray}
answered Jan 15 at 19:13
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Wow, thanks! As a follow-up, assuming that $x=frac{eta}{eta_c}$, how can I get rid of all $eta$'s and just express them as x?
$endgroup$
– Jackson Hart
Jan 15 at 20:16
$begingroup$
The first term becomes $$ frac{3}{4}frac{1}{eta_c^2(1 - eta/eta_c)^2} = frac{3}{4eta_c^2} frac{1}{(1 - x)^2} $$ the last one is the same except replace the factor in parenthesis by $x$
$endgroup$
– caverac
Jan 15 at 20:24
$begingroup$
Did you accidentally lose a $eta$? The last term in first and second lines. In first line, there is an $eta$ in denominator. Where is it on next line? After the $d eta$ shouldnt there also be an $eta$ in denominator? It looks like you dropped it?
$endgroup$
– Jackson Hart
Jan 15 at 20:48
$begingroup$
@JacksonHart Yes I did! Sorry about that. Fixed
$endgroup$
– caverac
Jan 15 at 20:52
add a comment |
$begingroup$
begin{eqnarray}
require{cancel}
frac{{rm d}}{{rm d}eta}g(1, eta) &=& frac{1}{4}frac{{rm d}}{{rm d}eta} left[ frac{3cancel{eta}}{cancel{eta}(eta_c - eta)} + sum_{k=1}^4 k A_k frac{eta^k}{eta eta_c^k}right] \
&=& frac{3}{4} frac{{rm d}}{{rm d}eta} frac{1}{eta_c - eta} + frac{1}{4}sum_{k=1}^4 frac{k A_k}{eta_c^k} frac{{rm d}eta^{k-1}}{{rm d}eta} \
&=& frac{3}{4}frac{1}{(eta_c - eta)^2} + frac{1}{4}sum_{k=1}^4frac{k(k-1) A_k}{eta_c^k}eta^{k - 2} \
&=& frac{3}{4}frac{1}{(eta_c - eta)^2} + frac{1}{4}sum_{k=1}^4frac{k(k-1) A_k}{eta_c^2}left(frac{eta}{eta_c}right)^{k - 2}
end{eqnarray}
answered Jan 15 at 19:13
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Wow, thanks! As a follow-up, assuming that $x=frac{eta}{eta_c}$, how can I get rid of all $eta$'s and just express them as x?
$endgroup$
– Jackson Hart
Jan 15 at 20:16
$begingroup$
The first term becomes $$ frac{3}{4}frac{1}{eta_c^2(1 - eta/eta_c)^2} = frac{3}{4eta_c^2} frac{1}{(1 - x)^2} $$ the last one is the same except replace the factor in parenthesis by $x$
$endgroup$
– caverac
Jan 15 at 20:24
$begingroup$
Did you accidentally lose a $eta$? The last term in first and second lines. In first line, there is an $eta$ in denominator. Where is it on next line? After the $d eta$ shouldnt there also be an $eta$ in denominator? It looks like you dropped it?
$endgroup$
– Jackson Hart
Jan 15 at 20:48
$begingroup$
@JacksonHart Yes I did! Sorry about that. Fixed
$endgroup$
– caverac
Jan 15 at 20:52
add a comment |
$begingroup$
begin{eqnarray}
require{cancel}
frac{{rm d}}{{rm d}eta}g(1, eta) &=& frac{1}{4}frac{{rm d}}{{rm d}eta} left[ frac{3cancel{eta}}{cancel{eta}(eta_c - eta)} + sum_{k=1}^4 k A_k frac{eta^k}{eta eta_c^k}right] \
&=& frac{3}{4} frac{{rm d}}{{rm d}eta} frac{1}{eta_c - eta} + frac{1}{4}sum_{k=1}^4 frac{k A_k}{eta_c^k} frac{{rm d}eta^{k-1}}{{rm d}eta} \
&=& frac{3}{4}frac{1}{(eta_c - eta)^2} + frac{1}{4}sum_{k=1}^4frac{k(k-1) A_k}{eta_c^k}eta^{k - 2} \
&=& frac{3}{4}frac{1}{(eta_c - eta)^2} + frac{1}{4}sum_{k=1}^4frac{k(k-1) A_k}{eta_c^2}left(frac{eta}{eta_c}right)^{k - 2}
end{eqnarray}
answered Jan 15 at 19:13
caveraccaverac
14.6k31130
$endgroup$
begin{eqnarray}
require{cancel}
frac{{rm d}}{{rm d}eta}g(1, eta) &=& frac{1}{4}frac{{rm d}}{{rm d}eta} left[ frac{3cancel{eta}}{cancel{eta}(eta_c - eta)} + sum_{k=1}^4 k A_k frac{eta^k}{eta eta_c^k}right] \
&=& frac{3}{4} frac{{rm d}}{{rm d}eta} frac{1}{eta_c - eta} + frac{1}{4}sum_{k=1}^4 frac{k A_k}{eta_c^k} frac{{rm d}eta^{k-1}}{{rm d}eta} \
&=& frac{3}{4}frac{1}{(eta_c - eta)^2} + frac{1}{4}sum_{k=1}^4frac{k(k-1) A_k}{eta_c^k}eta^{k - 2} \
&=& frac{3}{4}frac{1}{(eta_c - eta)^2} + frac{1}{4}sum_{k=1}^4frac{k(k-1) A_k}{eta_c^2}left(frac{eta}{eta_c}right)^{k - 2}
end{eqnarray}
answered Jan 15 at 19:13
caveraccaverac
14.6k31130
edited Jan 15 at 20:51
answered Jan 15 at 19:13
caveraccaverac
14.6k31130
answered Jan 15 at 19:13
caveraccaverac
14.6k31130
answered Jan 15 at 19:13
caveraccaverac
14.6k31130
14.6k31130
$begingroup$
Wow, thanks! As a follow-up, assuming that $x=frac{eta}{eta_c}$, how can I get rid of all $eta$'s and just express them as x?
$endgroup$
– Jackson Hart
Jan 15 at 20:16
$begingroup$
The first term becomes $$ frac{3}{4}frac{1}{eta_c^2(1 - eta/eta_c)^2} = frac{3}{4eta_c^2} frac{1}{(1 - x)^2} $$ the last one is the same except replace the factor in parenthesis by $x$
$endgroup$
– caverac
Jan 15 at 20:24
$begingroup$
Did you accidentally lose a $eta$? The last term in first and second lines. In first line, there is an $eta$ in denominator. Where is it on next line? After the $d eta$ shouldnt there also be an $eta$ in denominator? It looks like you dropped it?
$endgroup$
– Jackson Hart
Jan 15 at 20:48
$begingroup$
@JacksonHart Yes I did! Sorry about that. Fixed
$endgroup$
– caverac
Jan 15 at 20:52
add a comment |
$begingroup$
Wow, thanks! As a follow-up, assuming that $x=frac{eta}{eta_c}$, how can I get rid of all $eta$'s and just express them as x?
$endgroup$
– Jackson Hart
Jan 15 at 20:16
$begingroup$
The first term becomes $$ frac{3}{4}frac{1}{eta_c^2(1 - eta/eta_c)^2} = frac{3}{4eta_c^2} frac{1}{(1 - x)^2} $$ the last one is the same except replace the factor in parenthesis by $x$
$endgroup$
– caverac
Jan 15 at 20:24
$begingroup$
Did you accidentally lose a $eta$? The last term in first and second lines. In first line, there is an $eta$ in denominator. Where is it on next line? After the $d eta$ shouldnt there also be an $eta$ in denominator? It looks like you dropped it?
$endgroup$
– Jackson Hart
Jan 15 at 20:48
$begingroup$
@JacksonHart Yes I did! Sorry about that. Fixed
$endgroup$
– caverac
Jan 15 at 20:52
$begingroup$
Wow, thanks! As a follow-up, assuming that $x=frac{eta}{eta_c}$, how can I get rid of all $eta$'s and just express them as x?
$endgroup$
– Jackson Hart
Jan 15 at 20:16
$begingroup$
Wow, thanks! As a follow-up, assuming that $x=frac{eta}{eta_c}$, how can I get rid of all $eta$'s and just express them as x?
$endgroup$
– Jackson Hart
Jan 15 at 20:16
$begingroup$
The first term becomes $$ frac{3}{4}frac{1}{eta_c^2(1 - eta/eta_c)^2} = frac{3}{4eta_c^2} frac{1}{(1 - x)^2} $$ the last one is the same except replace the factor in parenthesis by $x$
$endgroup$
– caverac
Jan 15 at 20:24
$begingroup$
The first term becomes $$ frac{3}{4}frac{1}{eta_c^2(1 - eta/eta_c)^2} = frac{3}{4eta_c^2} frac{1}{(1 - x)^2} $$ the last one is the same except replace the factor in parenthesis by $x$
$endgroup$
– caverac
Jan 15 at 20:24
$begingroup$
Did you accidentally lose a $eta$? The last term in first and second lines. In first line, there is an $eta$ in denominator. Where is it on next line? After the $d eta$ shouldnt there also be an $eta$ in denominator? It looks like you dropped it?
$endgroup$
– Jackson Hart
Jan 15 at 20:48
$begingroup$
Did you accidentally lose a $eta$? The last term in first and second lines. In first line, there is an $eta$ in denominator. Where is it on next line? After the $d eta$ shouldnt there also be an $eta$ in denominator? It looks like you dropped it?
$endgroup$
– Jackson Hart
Jan 15 at 20:48
$begingroup$
@JacksonHart Yes I did! Sorry about that. Fixed
$endgroup$
– caverac
Jan 15 at 20:52
$begingroup$
@JacksonHart Yes I did! Sorry about that. Fixed
$endgroup$
– caverac
Jan 15 at 20:52
add a comment |
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