Mixing
Deriving $dy/dx = 2cos x/cos y$ given $sin y=2sin x$
$begingroup$
My original question is to find the second derivative of $sin y=2sin x$
I derived it once got $2cos x/cos y$ which was correct but the second time did not get $3sec^2ytan y$ which is the answer.
Im not sure where I've gone wrong, please help.
Thank you.
trigonometry implicit-differentiation
edited Jan 31 at 19:35
Larry
2,53031131
asked Jan 31 at 19:32
user639649user639649
527
$endgroup$
add a comment |
$begingroup$
My original question is to find the second derivative of $sin y=2sin x$
I derived it once got $2cos x/cos y$ which was correct but the second time did not get $3sec^2ytan y$ which is the answer.
Im not sure where I've gone wrong, please help.
Thank you.
trigonometry implicit-differentiation
edited Jan 31 at 19:35
Larry
2,53031131
asked Jan 31 at 19:32
user639649user639649
527
$endgroup$
add a comment |
$begingroup$
My original question is to find the second derivative of $sin y=2sin x$
I derived it once got $2cos x/cos y$ which was correct but the second time did not get $3sec^2ytan y$ which is the answer.
Im not sure where I've gone wrong, please help.
Thank you.
trigonometry implicit-differentiation
edited Jan 31 at 19:35
Larry
2,53031131
asked Jan 31 at 19:32
user639649user639649
527
$endgroup$
My original question is to find the second derivative of $sin y=2sin x$
I derived it once got $2cos x/cos y$ which was correct but the second time did not get $3sec^2ytan y$ which is the answer.
Im not sure where I've gone wrong, please help.
Thank you.
trigonometry implicit-differentiation
trigonometry implicit-differentiation
edited Jan 31 at 19:35
Larry
2,53031131
asked Jan 31 at 19:32
user639649user639649
527
edited Jan 31 at 19:35
Larry
2,53031131
asked Jan 31 at 19:32
user639649user639649
527
edited Jan 31 at 19:35
Larry
2,53031131
edited Jan 31 at 19:35
Larry
2,53031131
edited Jan 31 at 19:35
Larry
2,53031131
2,53031131
asked Jan 31 at 19:32
user639649user639649
527
asked Jan 31 at 19:32
user639649user639649
527
asked Jan 31 at 19:32
user639649user639649
527
527
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You got
$$frac{dy}{dx}=2frac{cos(x)}{cos(y)}$$
Recall the quotient rule, then the second derivative will be
$$begin{align}
frac{d^2y}{dx^2}&=2frac{d}{dx}frac{cos(x)}{cos(y)}\
&=2frac{-sin(x)cos(y)+sin(y)frac{dy}{dx}cos(x)}{cos^2(y)}\
&=frac{-sin(y)cos(y)+4tan(y)cos^2(x)}{cos^2(y)}
end{align}$$
We know that
$$sin(y)=2sin(x)$$
Then
$$sqrt{1-cos^2(y)}=2sqrt{1-cos^2(x)}$$
$$1-cos^2(y)=4(1-cos^2(x))$$
$$cos^2(x)=frac{3+cos^2(y)}{4}$$
Then
$$begin{align}
frac{d^2y}{dx^2}&=frac{-sin(y)cos(y)+tan(y)(3+cos^2(y))}{cos^2(y)}\
&=frac{-sin(y)cos(y)+3tan(y)+sin(x)cos(y)}{cos^2(y)}\
&=3sec^2(y)tan(y)
end{align}$$
answered Jan 31 at 20:01
LarryLarry
2,53031131
$endgroup$
$begingroup$
Ahh okay! This makes a lot of sense, i was wondering that when you manipulated the identity how you knew which root to take? Like why did you do the positive root instead of the negative?
$endgroup$
– user639649
Jan 31 at 20:05
$begingroup$
I am not sure if I totally understand your question. It doesn't really matter whether I take positive or negative root. When I square both sides, they will become positive.
$endgroup$
– Larry
Jan 31 at 20:25
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You got
$$frac{dy}{dx}=2frac{cos(x)}{cos(y)}$$
Recall the quotient rule, then the second derivative will be
$$begin{align}
frac{d^2y}{dx^2}&=2frac{d}{dx}frac{cos(x)}{cos(y)}\
&=2frac{-sin(x)cos(y)+sin(y)frac{dy}{dx}cos(x)}{cos^2(y)}\
&=frac{-sin(y)cos(y)+4tan(y)cos^2(x)}{cos^2(y)}
end{align}$$
We know that
$$sin(y)=2sin(x)$$
Then
$$sqrt{1-cos^2(y)}=2sqrt{1-cos^2(x)}$$
$$1-cos^2(y)=4(1-cos^2(x))$$
$$cos^2(x)=frac{3+cos^2(y)}{4}$$
Then
$$begin{align}
frac{d^2y}{dx^2}&=frac{-sin(y)cos(y)+tan(y)(3+cos^2(y))}{cos^2(y)}\
&=frac{-sin(y)cos(y)+3tan(y)+sin(x)cos(y)}{cos^2(y)}\
&=3sec^2(y)tan(y)
end{align}$$
answered Jan 31 at 20:01
LarryLarry
2,53031131
$endgroup$
$begingroup$
Ahh okay! This makes a lot of sense, i was wondering that when you manipulated the identity how you knew which root to take? Like why did you do the positive root instead of the negative?
$endgroup$
– user639649
Jan 31 at 20:05
$begingroup$
I am not sure if I totally understand your question. It doesn't really matter whether I take positive or negative root. When I square both sides, they will become positive.
$endgroup$
– Larry
Jan 31 at 20:25
add a comment |
$begingroup$
You got
$$frac{dy}{dx}=2frac{cos(x)}{cos(y)}$$
Recall the quotient rule, then the second derivative will be
$$begin{align}
frac{d^2y}{dx^2}&=2frac{d}{dx}frac{cos(x)}{cos(y)}\
&=2frac{-sin(x)cos(y)+sin(y)frac{dy}{dx}cos(x)}{cos^2(y)}\
&=frac{-sin(y)cos(y)+4tan(y)cos^2(x)}{cos^2(y)}
end{align}$$
We know that
$$sin(y)=2sin(x)$$
Then
$$sqrt{1-cos^2(y)}=2sqrt{1-cos^2(x)}$$
$$1-cos^2(y)=4(1-cos^2(x))$$
$$cos^2(x)=frac{3+cos^2(y)}{4}$$
Then
$$begin{align}
frac{d^2y}{dx^2}&=frac{-sin(y)cos(y)+tan(y)(3+cos^2(y))}{cos^2(y)}\
&=frac{-sin(y)cos(y)+3tan(y)+sin(x)cos(y)}{cos^2(y)}\
&=3sec^2(y)tan(y)
end{align}$$
answered Jan 31 at 20:01
LarryLarry
2,53031131
$endgroup$
$begingroup$
Ahh okay! This makes a lot of sense, i was wondering that when you manipulated the identity how you knew which root to take? Like why did you do the positive root instead of the negative?
$endgroup$
– user639649
Jan 31 at 20:05
$begingroup$
I am not sure if I totally understand your question. It doesn't really matter whether I take positive or negative root. When I square both sides, they will become positive.
$endgroup$
– Larry
Jan 31 at 20:25
add a comment |
$begingroup$
You got
$$frac{dy}{dx}=2frac{cos(x)}{cos(y)}$$
Recall the quotient rule, then the second derivative will be
$$begin{align}
frac{d^2y}{dx^2}&=2frac{d}{dx}frac{cos(x)}{cos(y)}\
&=2frac{-sin(x)cos(y)+sin(y)frac{dy}{dx}cos(x)}{cos^2(y)}\
&=frac{-sin(y)cos(y)+4tan(y)cos^2(x)}{cos^2(y)}
end{align}$$
We know that
$$sin(y)=2sin(x)$$
Then
$$sqrt{1-cos^2(y)}=2sqrt{1-cos^2(x)}$$
$$1-cos^2(y)=4(1-cos^2(x))$$
$$cos^2(x)=frac{3+cos^2(y)}{4}$$
Then
$$begin{align}
frac{d^2y}{dx^2}&=frac{-sin(y)cos(y)+tan(y)(3+cos^2(y))}{cos^2(y)}\
&=frac{-sin(y)cos(y)+3tan(y)+sin(x)cos(y)}{cos^2(y)}\
&=3sec^2(y)tan(y)
end{align}$$
answered Jan 31 at 20:01
LarryLarry
2,53031131
$endgroup$
You got
$$frac{dy}{dx}=2frac{cos(x)}{cos(y)}$$
Recall the quotient rule, then the second derivative will be
$$begin{align}
frac{d^2y}{dx^2}&=2frac{d}{dx}frac{cos(x)}{cos(y)}\
&=2frac{-sin(x)cos(y)+sin(y)frac{dy}{dx}cos(x)}{cos^2(y)}\
&=frac{-sin(y)cos(y)+4tan(y)cos^2(x)}{cos^2(y)}
end{align}$$
We know that
$$sin(y)=2sin(x)$$
Then
$$sqrt{1-cos^2(y)}=2sqrt{1-cos^2(x)}$$
$$1-cos^2(y)=4(1-cos^2(x))$$
$$cos^2(x)=frac{3+cos^2(y)}{4}$$
Then
$$begin{align}
frac{d^2y}{dx^2}&=frac{-sin(y)cos(y)+tan(y)(3+cos^2(y))}{cos^2(y)}\
&=frac{-sin(y)cos(y)+3tan(y)+sin(x)cos(y)}{cos^2(y)}\
&=3sec^2(y)tan(y)
end{align}$$
answered Jan 31 at 20:01
LarryLarry
2,53031131
answered Jan 31 at 20:01
LarryLarry
2,53031131
answered Jan 31 at 20:01
LarryLarry
2,53031131
answered Jan 31 at 20:01
LarryLarry
2,53031131
2,53031131
$begingroup$
Ahh okay! This makes a lot of sense, i was wondering that when you manipulated the identity how you knew which root to take? Like why did you do the positive root instead of the negative?
$endgroup$
– user639649
Jan 31 at 20:05
$begingroup$
I am not sure if I totally understand your question. It doesn't really matter whether I take positive or negative root. When I square both sides, they will become positive.
$endgroup$
– Larry
Jan 31 at 20:25
add a comment |
$begingroup$
Ahh okay! This makes a lot of sense, i was wondering that when you manipulated the identity how you knew which root to take? Like why did you do the positive root instead of the negative?
$endgroup$
– user639649
Jan 31 at 20:05
$begingroup$
I am not sure if I totally understand your question. It doesn't really matter whether I take positive or negative root. When I square both sides, they will become positive.
$endgroup$
– Larry
Jan 31 at 20:25
$begingroup$
Ahh okay! This makes a lot of sense, i was wondering that when you manipulated the identity how you knew which root to take? Like why did you do the positive root instead of the negative?
$endgroup$
– user639649
Jan 31 at 20:05
$begingroup$
Ahh okay! This makes a lot of sense, i was wondering that when you manipulated the identity how you knew which root to take? Like why did you do the positive root instead of the negative?
$endgroup$
– user639649
Jan 31 at 20:05
$begingroup$
I am not sure if I totally understand your question. It doesn't really matter whether I take positive or negative root. When I square both sides, they will become positive.
$endgroup$
– Larry
Jan 31 at 20:25
$begingroup$
I am not sure if I totally understand your question. It doesn't really matter whether I take positive or negative root. When I square both sides, they will become positive.
$endgroup$
– Larry
Jan 31 at 20:25
add a comment |
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