Hai… my problem is i dont know how to solve this question because I confuse to use any method
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$e^{n+y}=1+5sqrt{y^2}-lnleft[frac{y^5(6n^2+1)}{sqrt{2n^3-4}}right]$
Find $frac{dy}{dn}$ when $n=0$ and $y=1$.
How can I approach this problem?
(original link: https://i.stack.imgur.com/W9Mym.jpg)
differential-equations implicit-differentiation
edited Nov 16 at 17:06
Nathaniel D. Hoffman
357
asked Nov 16 at 16:51
Farhan Fadzlie
4
add a comment |
up vote
-3
down vote
favorite
$e^{n+y}=1+5sqrt{y^2}-lnleft[frac{y^5(6n^2+1)}{sqrt{2n^3-4}}right]$
Find $frac{dy}{dn}$ when $n=0$ and $y=1$.
How can I approach this problem?
(original link: https://i.stack.imgur.com/W9Mym.jpg)
differential-equations implicit-differentiation
edited Nov 16 at 17:06
Nathaniel D. Hoffman
357
asked Nov 16 at 16:51
Farhan Fadzlie
4
2
Something there looks like $;n;$ , something else like $;x;$ (with some imagination), something like $;ln;$ ...or maybe $;in;$ ...Try to type your question, please.
– DonAntonio
Nov 16 at 16:54
Hi, and welcome to Math.SE. Please use Math Jax to type your questions, in order to make the mathematical text comprehensible to all members.
– Daniele Tampieri
Nov 16 at 17:01
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
$e^{n+y}=1+5sqrt{y^2}-lnleft[frac{y^5(6n^2+1)}{sqrt{2n^3-4}}right]$
Find $frac{dy}{dn}$ when $n=0$ and $y=1$.
How can I approach this problem?
(original link: https://i.stack.imgur.com/W9Mym.jpg)
differential-equations implicit-differentiation
edited Nov 16 at 17:06
Nathaniel D. Hoffman
357
asked Nov 16 at 16:51
Farhan Fadzlie
4
$e^{n+y}=1+5sqrt{y^2}-lnleft[frac{y^5(6n^2+1)}{sqrt{2n^3-4}}right]$
Find $frac{dy}{dn}$ when $n=0$ and $y=1$.
How can I approach this problem?
(original link: https://i.stack.imgur.com/W9Mym.jpg)
differential-equations implicit-differentiation
differential-equations implicit-differentiation
edited Nov 16 at 17:06
Nathaniel D. Hoffman
357
asked Nov 16 at 16:51
Farhan Fadzlie
4
edited Nov 16 at 17:06
Nathaniel D. Hoffman
357
asked Nov 16 at 16:51
Farhan Fadzlie
4
edited Nov 16 at 17:06
Nathaniel D. Hoffman
357
edited Nov 16 at 17:06
Nathaniel D. Hoffman
357
edited Nov 16 at 17:06
Nathaniel D. Hoffman
357
357
asked Nov 16 at 16:51
Farhan Fadzlie
4
asked Nov 16 at 16:51
Farhan Fadzlie
4
asked Nov 16 at 16:51
Farhan Fadzlie
4
4
2
Something there looks like $;n;$ , something else like $;x;$ (with some imagination), something like $;ln;$ ...or maybe $;in;$ ...Try to type your question, please.
– DonAntonio
Nov 16 at 16:54
Hi, and welcome to Math.SE. Please use Math Jax to type your questions, in order to make the mathematical text comprehensible to all members.
– Daniele Tampieri
Nov 16 at 17:01
add a comment |
2
Something there looks like $;n;$ , something else like $;x;$ (with some imagination), something like $;ln;$ ...or maybe $;in;$ ...Try to type your question, please.
– DonAntonio
Nov 16 at 16:54
Hi, and welcome to Math.SE. Please use Math Jax to type your questions, in order to make the mathematical text comprehensible to all members.
– Daniele Tampieri
Nov 16 at 17:01
2
2
Something there looks like $;n;$ , something else like $;x;$ (with some imagination), something like $;ln;$ ...or maybe $;in;$ ...Try to type your question, please.
– DonAntonio
Nov 16 at 16:54
Something there looks like $;n;$ , something else like $;x;$ (with some imagination), something like $;ln;$ ...or maybe $;in;$ ...Try to type your question, please.
– DonAntonio
Nov 16 at 16:54
Hi, and welcome to Math.SE. Please use Math Jax to type your questions, in order to make the mathematical text comprehensible to all members.
– Daniele Tampieri
Nov 16 at 17:01
Hi, and welcome to Math.SE. Please use Math Jax to type your questions, in order to make the mathematical text comprehensible to all members.
– Daniele Tampieri
Nov 16 at 17:01
add a comment |
1 Answer
1
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0
down vote
When $n=0$ and $y=1$, the equation reads: $e = 1+5-lnleft[-frac{1}{2imath}right]$, so $e$ has been equated to a complex number (in fact there are no solutions for this equation on the $n=0$ line). If you meant to write "$imath n$" instead of $ln$, then the equation still reads $e=6$. Please clarify your question, and I will edit my answer accordingly.
answered Nov 16 at 17:12
Nathaniel D. Hoffman
357
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
When $n=0$ and $y=1$, the equation reads: $e = 1+5-lnleft[-frac{1}{2imath}right]$, so $e$ has been equated to a complex number (in fact there are no solutions for this equation on the $n=0$ line). If you meant to write "$imath n$" instead of $ln$, then the equation still reads $e=6$. Please clarify your question, and I will edit my answer accordingly.
answered Nov 16 at 17:12
Nathaniel D. Hoffman
357
add a comment |
up vote
0
down vote
When $n=0$ and $y=1$, the equation reads: $e = 1+5-lnleft[-frac{1}{2imath}right]$, so $e$ has been equated to a complex number (in fact there are no solutions for this equation on the $n=0$ line). If you meant to write "$imath n$" instead of $ln$, then the equation still reads $e=6$. Please clarify your question, and I will edit my answer accordingly.
answered Nov 16 at 17:12
Nathaniel D. Hoffman
357
add a comment |
up vote
0
down vote
up vote
0
down vote
When $n=0$ and $y=1$, the equation reads: $e = 1+5-lnleft[-frac{1}{2imath}right]$, so $e$ has been equated to a complex number (in fact there are no solutions for this equation on the $n=0$ line). If you meant to write "$imath n$" instead of $ln$, then the equation still reads $e=6$. Please clarify your question, and I will edit my answer accordingly.
answered Nov 16 at 17:12
Nathaniel D. Hoffman
357
When $n=0$ and $y=1$, the equation reads: $e = 1+5-lnleft[-frac{1}{2imath}right]$, so $e$ has been equated to a complex number (in fact there are no solutions for this equation on the $n=0$ line). If you meant to write "$imath n$" instead of $ln$, then the equation still reads $e=6$. Please clarify your question, and I will edit my answer accordingly.
answered Nov 16 at 17:12
Nathaniel D. Hoffman
357
answered Nov 16 at 17:12
Nathaniel D. Hoffman
357
answered Nov 16 at 17:12
Nathaniel D. Hoffman
357
answered Nov 16 at 17:12
Nathaniel D. Hoffman
357
357
add a comment |
add a comment |
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Something there looks like $;n;$ , something else like $;x;$ (with some imagination), something like $;ln;$ ...or maybe $;in;$ ...Try to type your question, please.
– DonAntonio
Nov 16 at 16:54
Hi, and welcome to Math.SE. Please use Math Jax to type your questions, in order to make the mathematical text comprehensible to all members.
– Daniele Tampieri
Nov 16 at 17:01