Mixing
How to find the second derivative(with respect to x) of cos y + sin y = x?
$begingroup$
The answer is $$pm dfrac {x}{(2-x^2)^{3/2}}$$
I dont understand how it goes from using just $cos x$ and $sin x$ to simple terms and a root. I've tried multiple times and have not gotten anywhere near the answer.
Thanks.
derivatives trigonometry
edited Jan 30 at 0:01
bjcolby15
1,51711016
asked Jan 29 at 22:49
user639649user639649
416
$endgroup$
add a comment |
$begingroup$
The answer is $$pm dfrac {x}{(2-x^2)^{3/2}}$$
I dont understand how it goes from using just $cos x$ and $sin x$ to simple terms and a root. I've tried multiple times and have not gotten anywhere near the answer.
Thanks.
derivatives trigonometry
edited Jan 30 at 0:01
bjcolby15
1,51711016
asked Jan 29 at 22:49
user639649user639649
416
$endgroup$
add a comment |
$begingroup$
The answer is $$pm dfrac {x}{(2-x^2)^{3/2}}$$
I dont understand how it goes from using just $cos x$ and $sin x$ to simple terms and a root. I've tried multiple times and have not gotten anywhere near the answer.
Thanks.
derivatives trigonometry
edited Jan 30 at 0:01
bjcolby15
1,51711016
asked Jan 29 at 22:49
user639649user639649
416
$endgroup$
The answer is $$pm dfrac {x}{(2-x^2)^{3/2}}$$
I dont understand how it goes from using just $cos x$ and $sin x$ to simple terms and a root. I've tried multiple times and have not gotten anywhere near the answer.
Thanks.
derivatives trigonometry
derivatives trigonometry
edited Jan 30 at 0:01
bjcolby15
1,51711016
asked Jan 29 at 22:49
user639649user639649
416
edited Jan 30 at 0:01
bjcolby15
1,51711016
asked Jan 29 at 22:49
user639649user639649
416
edited Jan 30 at 0:01
bjcolby15
1,51711016
edited Jan 30 at 0:01
bjcolby15
1,51711016
edited Jan 30 at 0:01
bjcolby15
1,51711016
1,51711016
asked Jan 29 at 22:49
user639649user639649
416
asked Jan 29 at 22:49
user639649user639649
416
asked Jan 29 at 22:49
user639649user639649
416
416
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In this case you can write $y$ explicitly in terms of $x$... up to a choice of branch for the $arcsin$ function.
$$
sin y + cos y = x Rightarrow (sin y + cos y)^2 = x ^2 Leftrightarrow \
sin^2 y + 2 sin y cos y + cos^2 y = x^2 Leftrightarrow\
sin(2y) = x^2-1 Leftrightarrow \
y = frac 12 arcsin(x^2-1)
$$
Hence,
$$
y'(x) = pm frac 12 frac{2x}{sqrt{1-(x^2-1)^2}}=pm x (1-(x^2-1)^2)^{-1/2}
$$
Can you proceed?
Note that the relation does not globally define $y$ as a function of $x$... This is why we have the $pm$, the "graph" would be a periodic function about the $y$ axis and when you pick a branch the function can be increasing or decreasing.
answered Jan 29 at 23:48
PierreCarrePierreCarre
1,665212
$endgroup$
$begingroup$
Ahh okay it makes sense, so does this mean that whenever you're being asked to derive a question that involves both x,y and cos and sin you should try and opt to write the equation explicitly and from then derive?
$endgroup$
– user639649
Jan 30 at 0:03
$begingroup$
Sure, you can try to write y as an explicit function of x. If it works, fine, but it will normally not possible. In general, you should substitute y=y(x) in the equation, derive both sides and solve for y'(x). You will not get an explicit formula for y'(x) but a formula envolving x and y(x).
$endgroup$
– PierreCarre
Jan 30 at 8:13
add a comment |
$begingroup$
If you square both sides of the equation you get
$$sin y+cos y=ximpliessin (2y)=x^2-1implies y=frac12arcsin(x^2-1)$$
Now differentiate the last expression and find if you are missing some solution of the original equation. Note that $xin[-sqrt 2,sqrt 2]$.
answered Jan 29 at 22:59
MasacrosoMasacroso
13.1k41747
$endgroup$
$begingroup$
So I got the answer! But I only got a negative value, why is it +/- ? Because of the domain ? Also I was wondering how you knew to use the method that you had used, since it hadn't occurred to me to use that same method
$endgroup$
– user639649
Jan 29 at 23:16
$begingroup$
@user639649 when you square you probably lose some set of solutions. I tried to simplify the LHS, and squaring it have a nice expression, this is why I did it
$endgroup$
– Masacroso
Jan 29 at 23:21
$begingroup$
@user639649 also with the arc sine function you are reducing the solutions of the original equation
$endgroup$
– Masacroso
Jan 29 at 23:28
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In this case you can write $y$ explicitly in terms of $x$... up to a choice of branch for the $arcsin$ function.
$$
sin y + cos y = x Rightarrow (sin y + cos y)^2 = x ^2 Leftrightarrow \
sin^2 y + 2 sin y cos y + cos^2 y = x^2 Leftrightarrow\
sin(2y) = x^2-1 Leftrightarrow \
y = frac 12 arcsin(x^2-1)
$$
Hence,
$$
y'(x) = pm frac 12 frac{2x}{sqrt{1-(x^2-1)^2}}=pm x (1-(x^2-1)^2)^{-1/2}
$$
Can you proceed?
Note that the relation does not globally define $y$ as a function of $x$... This is why we have the $pm$, the "graph" would be a periodic function about the $y$ axis and when you pick a branch the function can be increasing or decreasing.
answered Jan 29 at 23:48
PierreCarrePierreCarre
1,665212
$endgroup$
$begingroup$
Ahh okay it makes sense, so does this mean that whenever you're being asked to derive a question that involves both x,y and cos and sin you should try and opt to write the equation explicitly and from then derive?
$endgroup$
– user639649
Jan 30 at 0:03
$begingroup$
Sure, you can try to write y as an explicit function of x. If it works, fine, but it will normally not possible. In general, you should substitute y=y(x) in the equation, derive both sides and solve for y'(x). You will not get an explicit formula for y'(x) but a formula envolving x and y(x).
$endgroup$
– PierreCarre
Jan 30 at 8:13
add a comment |
$begingroup$
In this case you can write $y$ explicitly in terms of $x$... up to a choice of branch for the $arcsin$ function.
$$
sin y + cos y = x Rightarrow (sin y + cos y)^2 = x ^2 Leftrightarrow \
sin^2 y + 2 sin y cos y + cos^2 y = x^2 Leftrightarrow\
sin(2y) = x^2-1 Leftrightarrow \
y = frac 12 arcsin(x^2-1)
$$
Hence,
$$
y'(x) = pm frac 12 frac{2x}{sqrt{1-(x^2-1)^2}}=pm x (1-(x^2-1)^2)^{-1/2}
$$
Can you proceed?
Note that the relation does not globally define $y$ as a function of $x$... This is why we have the $pm$, the "graph" would be a periodic function about the $y$ axis and when you pick a branch the function can be increasing or decreasing.
answered Jan 29 at 23:48
PierreCarrePierreCarre
1,665212
$endgroup$
$begingroup$
Ahh okay it makes sense, so does this mean that whenever you're being asked to derive a question that involves both x,y and cos and sin you should try and opt to write the equation explicitly and from then derive?
$endgroup$
– user639649
Jan 30 at 0:03
$begingroup$
Sure, you can try to write y as an explicit function of x. If it works, fine, but it will normally not possible. In general, you should substitute y=y(x) in the equation, derive both sides and solve for y'(x). You will not get an explicit formula for y'(x) but a formula envolving x and y(x).
$endgroup$
– PierreCarre
Jan 30 at 8:13
add a comment |
$begingroup$
In this case you can write $y$ explicitly in terms of $x$... up to a choice of branch for the $arcsin$ function.
$$
sin y + cos y = x Rightarrow (sin y + cos y)^2 = x ^2 Leftrightarrow \
sin^2 y + 2 sin y cos y + cos^2 y = x^2 Leftrightarrow\
sin(2y) = x^2-1 Leftrightarrow \
y = frac 12 arcsin(x^2-1)
$$
Hence,
$$
y'(x) = pm frac 12 frac{2x}{sqrt{1-(x^2-1)^2}}=pm x (1-(x^2-1)^2)^{-1/2}
$$
Can you proceed?
Note that the relation does not globally define $y$ as a function of $x$... This is why we have the $pm$, the "graph" would be a periodic function about the $y$ axis and when you pick a branch the function can be increasing or decreasing.
answered Jan 29 at 23:48
PierreCarrePierreCarre
1,665212
$endgroup$
In this case you can write $y$ explicitly in terms of $x$... up to a choice of branch for the $arcsin$ function.
$$
sin y + cos y = x Rightarrow (sin y + cos y)^2 = x ^2 Leftrightarrow \
sin^2 y + 2 sin y cos y + cos^2 y = x^2 Leftrightarrow\
sin(2y) = x^2-1 Leftrightarrow \
y = frac 12 arcsin(x^2-1)
$$
Hence,
$$
y'(x) = pm frac 12 frac{2x}{sqrt{1-(x^2-1)^2}}=pm x (1-(x^2-1)^2)^{-1/2}
$$
Can you proceed?
Note that the relation does not globally define $y$ as a function of $x$... This is why we have the $pm$, the "graph" would be a periodic function about the $y$ axis and when you pick a branch the function can be increasing or decreasing.
answered Jan 29 at 23:48
PierreCarrePierreCarre
1,665212
answered Jan 29 at 23:48
PierreCarrePierreCarre
1,665212
answered Jan 29 at 23:48
PierreCarrePierreCarre
1,665212
answered Jan 29 at 23:48
PierreCarrePierreCarre
1,665212
1,665212
$begingroup$
Ahh okay it makes sense, so does this mean that whenever you're being asked to derive a question that involves both x,y and cos and sin you should try and opt to write the equation explicitly and from then derive?
$endgroup$
– user639649
Jan 30 at 0:03
$begingroup$
Sure, you can try to write y as an explicit function of x. If it works, fine, but it will normally not possible. In general, you should substitute y=y(x) in the equation, derive both sides and solve for y'(x). You will not get an explicit formula for y'(x) but a formula envolving x and y(x).
$endgroup$
– PierreCarre
Jan 30 at 8:13
add a comment |
$begingroup$
Ahh okay it makes sense, so does this mean that whenever you're being asked to derive a question that involves both x,y and cos and sin you should try and opt to write the equation explicitly and from then derive?
$endgroup$
– user639649
Jan 30 at 0:03
$begingroup$
Sure, you can try to write y as an explicit function of x. If it works, fine, but it will normally not possible. In general, you should substitute y=y(x) in the equation, derive both sides and solve for y'(x). You will not get an explicit formula for y'(x) but a formula envolving x and y(x).
$endgroup$
– PierreCarre
Jan 30 at 8:13
$begingroup$
Ahh okay it makes sense, so does this mean that whenever you're being asked to derive a question that involves both x,y and cos and sin you should try and opt to write the equation explicitly and from then derive?
$endgroup$
– user639649
Jan 30 at 0:03
$begingroup$
Ahh okay it makes sense, so does this mean that whenever you're being asked to derive a question that involves both x,y and cos and sin you should try and opt to write the equation explicitly and from then derive?
$endgroup$
– user639649
Jan 30 at 0:03
$begingroup$
Sure, you can try to write y as an explicit function of x. If it works, fine, but it will normally not possible. In general, you should substitute y=y(x) in the equation, derive both sides and solve for y'(x). You will not get an explicit formula for y'(x) but a formula envolving x and y(x).
$endgroup$
– PierreCarre
Jan 30 at 8:13
$begingroup$
Sure, you can try to write y as an explicit function of x. If it works, fine, but it will normally not possible. In general, you should substitute y=y(x) in the equation, derive both sides and solve for y'(x). You will not get an explicit formula for y'(x) but a formula envolving x and y(x).
$endgroup$
– PierreCarre
Jan 30 at 8:13
add a comment |
$begingroup$
If you square both sides of the equation you get
$$sin y+cos y=ximpliessin (2y)=x^2-1implies y=frac12arcsin(x^2-1)$$
Now differentiate the last expression and find if you are missing some solution of the original equation. Note that $xin[-sqrt 2,sqrt 2]$.
answered Jan 29 at 22:59
MasacrosoMasacroso
13.1k41747
$endgroup$
$begingroup$
So I got the answer! But I only got a negative value, why is it +/- ? Because of the domain ? Also I was wondering how you knew to use the method that you had used, since it hadn't occurred to me to use that same method
$endgroup$
– user639649
Jan 29 at 23:16
$begingroup$
@user639649 when you square you probably lose some set of solutions. I tried to simplify the LHS, and squaring it have a nice expression, this is why I did it
$endgroup$
– Masacroso
Jan 29 at 23:21
$begingroup$
@user639649 also with the arc sine function you are reducing the solutions of the original equation
$endgroup$
– Masacroso
Jan 29 at 23:28
add a comment |
$begingroup$
If you square both sides of the equation you get
$$sin y+cos y=ximpliessin (2y)=x^2-1implies y=frac12arcsin(x^2-1)$$
Now differentiate the last expression and find if you are missing some solution of the original equation. Note that $xin[-sqrt 2,sqrt 2]$.
answered Jan 29 at 22:59
MasacrosoMasacroso
13.1k41747
$endgroup$
$begingroup$
So I got the answer! But I only got a negative value, why is it +/- ? Because of the domain ? Also I was wondering how you knew to use the method that you had used, since it hadn't occurred to me to use that same method
$endgroup$
– user639649
Jan 29 at 23:16
$begingroup$
@user639649 when you square you probably lose some set of solutions. I tried to simplify the LHS, and squaring it have a nice expression, this is why I did it
$endgroup$
– Masacroso
Jan 29 at 23:21
$begingroup$
@user639649 also with the arc sine function you are reducing the solutions of the original equation
$endgroup$
– Masacroso
Jan 29 at 23:28
add a comment |
$begingroup$
If you square both sides of the equation you get
$$sin y+cos y=ximpliessin (2y)=x^2-1implies y=frac12arcsin(x^2-1)$$
Now differentiate the last expression and find if you are missing some solution of the original equation. Note that $xin[-sqrt 2,sqrt 2]$.
answered Jan 29 at 22:59
MasacrosoMasacroso
13.1k41747
$endgroup$
If you square both sides of the equation you get
$$sin y+cos y=ximpliessin (2y)=x^2-1implies y=frac12arcsin(x^2-1)$$
Now differentiate the last expression and find if you are missing some solution of the original equation. Note that $xin[-sqrt 2,sqrt 2]$.
answered Jan 29 at 22:59
MasacrosoMasacroso
13.1k41747
answered Jan 29 at 22:59
MasacrosoMasacroso
13.1k41747
answered Jan 29 at 22:59
MasacrosoMasacroso
13.1k41747
answered Jan 29 at 22:59
MasacrosoMasacroso
13.1k41747
13.1k41747
$begingroup$
So I got the answer! But I only got a negative value, why is it +/- ? Because of the domain ? Also I was wondering how you knew to use the method that you had used, since it hadn't occurred to me to use that same method
$endgroup$
– user639649
Jan 29 at 23:16
$begingroup$
@user639649 when you square you probably lose some set of solutions. I tried to simplify the LHS, and squaring it have a nice expression, this is why I did it
$endgroup$
– Masacroso
Jan 29 at 23:21
$begingroup$
@user639649 also with the arc sine function you are reducing the solutions of the original equation
$endgroup$
– Masacroso
Jan 29 at 23:28
add a comment |
$begingroup$
So I got the answer! But I only got a negative value, why is it +/- ? Because of the domain ? Also I was wondering how you knew to use the method that you had used, since it hadn't occurred to me to use that same method
$endgroup$
– user639649
Jan 29 at 23:16
$begingroup$
@user639649 when you square you probably lose some set of solutions. I tried to simplify the LHS, and squaring it have a nice expression, this is why I did it
$endgroup$
– Masacroso
Jan 29 at 23:21
$begingroup$
@user639649 also with the arc sine function you are reducing the solutions of the original equation
$endgroup$
– Masacroso
Jan 29 at 23:28
$begingroup$
So I got the answer! But I only got a negative value, why is it +/- ? Because of the domain ? Also I was wondering how you knew to use the method that you had used, since it hadn't occurred to me to use that same method
$endgroup$
– user639649
Jan 29 at 23:16
$begingroup$
So I got the answer! But I only got a negative value, why is it +/- ? Because of the domain ? Also I was wondering how you knew to use the method that you had used, since it hadn't occurred to me to use that same method
$endgroup$
– user639649
Jan 29 at 23:16
$begingroup$
@user639649 when you square you probably lose some set of solutions. I tried to simplify the LHS, and squaring it have a nice expression, this is why I did it
$endgroup$
– Masacroso
Jan 29 at 23:21
$begingroup$
@user639649 when you square you probably lose some set of solutions. I tried to simplify the LHS, and squaring it have a nice expression, this is why I did it
$endgroup$
– Masacroso
Jan 29 at 23:21
$begingroup$
@user639649 also with the arc sine function you are reducing the solutions of the original equation
$endgroup$
– Masacroso
Jan 29 at 23:28
$begingroup$
@user639649 also with the arc sine function you are reducing the solutions of the original equation
$endgroup$
– Masacroso
Jan 29 at 23:28
add a comment |
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