Mixing
Implicit differentiation: $xcos (4x+3y)=ysin x$
$begingroup$
$$xcos (4x+3y)=ysin x$$
I have been stuck on this problem for the longest. I have the answer but I don't know how to get to it. I have used the product and chain rule on both sides. I keep getting this:
$$frac{cos(4x+3y)-4xsin(4x+3y)-ycos x}{sin x}$$
Here is the answer:
$$frac{mathrm dy}{mathrm dx} = frac{cos (4x+3y)-ycos x-4xsin (4x+3y)}{sin x + 3xsin (4x+3y)}$$
calculus trigonometry derivatives implicit-differentiation
edited Jan 28 at 9:40
Blue
49.3k870157
asked Mar 8 '14 at 5:24
Kelli DavisKelli Davis
2815
$endgroup$
add a comment |
$begingroup$
$$xcos (4x+3y)=ysin x$$
I have been stuck on this problem for the longest. I have the answer but I don't know how to get to it. I have used the product and chain rule on both sides. I keep getting this:
$$frac{cos(4x+3y)-4xsin(4x+3y)-ycos x}{sin x}$$
Here is the answer:
$$frac{mathrm dy}{mathrm dx} = frac{cos (4x+3y)-ycos x-4xsin (4x+3y)}{sin x + 3xsin (4x+3y)}$$
calculus trigonometry derivatives implicit-differentiation
edited Jan 28 at 9:40
Blue
49.3k870157
asked Mar 8 '14 at 5:24
Kelli DavisKelli Davis
2815
$endgroup$
$begingroup$
You'll have to use the product rule on each side.
$endgroup$
– user61527
Mar 8 '14 at 5:27
$begingroup$
Write $F(x,y)$ as ($lhs-rhs$). Compute the partial derivatives of $F(x,y)$ with respect to $x$ and then to $y$ and compute the classical ratio from total differentiation.
$endgroup$
– Claude Leibovici
Mar 8 '14 at 5:33
add a comment |
$begingroup$
$$xcos (4x+3y)=ysin x$$
I have been stuck on this problem for the longest. I have the answer but I don't know how to get to it. I have used the product and chain rule on both sides. I keep getting this:
$$frac{cos(4x+3y)-4xsin(4x+3y)-ycos x}{sin x}$$
Here is the answer:
$$frac{mathrm dy}{mathrm dx} = frac{cos (4x+3y)-ycos x-4xsin (4x+3y)}{sin x + 3xsin (4x+3y)}$$
calculus trigonometry derivatives implicit-differentiation
edited Jan 28 at 9:40
Blue
49.3k870157
asked Mar 8 '14 at 5:24
Kelli DavisKelli Davis
2815
$endgroup$
$$xcos (4x+3y)=ysin x$$
I have been stuck on this problem for the longest. I have the answer but I don't know how to get to it. I have used the product and chain rule on both sides. I keep getting this:
$$frac{cos(4x+3y)-4xsin(4x+3y)-ycos x}{sin x}$$
Here is the answer:
$$frac{mathrm dy}{mathrm dx} = frac{cos (4x+3y)-ycos x-4xsin (4x+3y)}{sin x + 3xsin (4x+3y)}$$
calculus trigonometry derivatives implicit-differentiation
calculus trigonometry derivatives implicit-differentiation
edited Jan 28 at 9:40
Blue
49.3k870157
asked Mar 8 '14 at 5:24
Kelli DavisKelli Davis
2815
edited Jan 28 at 9:40
Blue
49.3k870157
asked Mar 8 '14 at 5:24
Kelli DavisKelli Davis
2815
edited Jan 28 at 9:40
Blue
49.3k870157
edited Jan 28 at 9:40
Blue
49.3k870157
edited Jan 28 at 9:40
Blue
49.3k870157
49.3k870157
asked Mar 8 '14 at 5:24
Kelli DavisKelli Davis
2815
asked Mar 8 '14 at 5:24
Kelli DavisKelli Davis
2815
asked Mar 8 '14 at 5:24
Kelli DavisKelli Davis
2815
2815
$begingroup$
You'll have to use the product rule on each side.
$endgroup$
– user61527
Mar 8 '14 at 5:27
$begingroup$
Write $F(x,y)$ as ($lhs-rhs$). Compute the partial derivatives of $F(x,y)$ with respect to $x$ and then to $y$ and compute the classical ratio from total differentiation.
$endgroup$
– Claude Leibovici
Mar 8 '14 at 5:33
add a comment |
$begingroup$
You'll have to use the product rule on each side.
$endgroup$
– user61527
Mar 8 '14 at 5:27
$begingroup$
Write $F(x,y)$ as ($lhs-rhs$). Compute the partial derivatives of $F(x,y)$ with respect to $x$ and then to $y$ and compute the classical ratio from total differentiation.
$endgroup$
– Claude Leibovici
Mar 8 '14 at 5:33
$begingroup$
You'll have to use the product rule on each side.
$endgroup$
– user61527
Mar 8 '14 at 5:27
$begingroup$
You'll have to use the product rule on each side.
$endgroup$
– user61527
Mar 8 '14 at 5:27
$begingroup$
Write $F(x,y)$ as ($lhs-rhs$). Compute the partial derivatives of $F(x,y)$ with respect to $x$ and then to $y$ and compute the classical ratio from total differentiation.
$endgroup$
– Claude Leibovici
Mar 8 '14 at 5:33
$begingroup$
Write $F(x,y)$ as ($lhs-rhs$). Compute the partial derivatives of $F(x,y)$ with respect to $x$ and then to $y$ and compute the classical ratio from total differentiation.
$endgroup$
– Claude Leibovici
Mar 8 '14 at 5:33
add a comment |
1 Answer
1
active
oldest
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$begingroup$
$sin(4x+3y)+xcos(4x+3y)(4+3dy/dx)=(dy/dx)cos x -ysin x$
$sin(4x+3y)+4xcos(4x+3y)+3(dy/dx)cos(4x+3y)=(dy/dx)cos x -ysin x$
$sin(4x+3y)+4xcos(4x+3y)+ysin x=(dy/dx)cos x + 3(dy/dx)cos(4x+3y)$
$sin(4x+3y)+4xcos(4x+3y)+ysin x=(dy/dx)(cos x + 3cos(4x+3y))$
answered Mar 8 '14 at 5:35
kmitovkmitov
4,0002816
$endgroup$
$begingroup$
I don't see any $5y$ in the question --- was there an edit?
$endgroup$
– Gerry Myerson
Mar 8 '14 at 7:20
$begingroup$
Yes, it is my mkistake. It is 3. I corrected it.
$endgroup$
– kmitov
Mar 8 '14 at 7:52
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$sin(4x+3y)+xcos(4x+3y)(4+3dy/dx)=(dy/dx)cos x -ysin x$
$sin(4x+3y)+4xcos(4x+3y)+3(dy/dx)cos(4x+3y)=(dy/dx)cos x -ysin x$
$sin(4x+3y)+4xcos(4x+3y)+ysin x=(dy/dx)cos x + 3(dy/dx)cos(4x+3y)$
$sin(4x+3y)+4xcos(4x+3y)+ysin x=(dy/dx)(cos x + 3cos(4x+3y))$
answered Mar 8 '14 at 5:35
kmitovkmitov
4,0002816
$endgroup$
$begingroup$
I don't see any $5y$ in the question --- was there an edit?
$endgroup$
– Gerry Myerson
Mar 8 '14 at 7:20
$begingroup$
Yes, it is my mkistake. It is 3. I corrected it.
$endgroup$
– kmitov
Mar 8 '14 at 7:52
add a comment |
$begingroup$
$sin(4x+3y)+xcos(4x+3y)(4+3dy/dx)=(dy/dx)cos x -ysin x$
$sin(4x+3y)+4xcos(4x+3y)+3(dy/dx)cos(4x+3y)=(dy/dx)cos x -ysin x$
$sin(4x+3y)+4xcos(4x+3y)+ysin x=(dy/dx)cos x + 3(dy/dx)cos(4x+3y)$
$sin(4x+3y)+4xcos(4x+3y)+ysin x=(dy/dx)(cos x + 3cos(4x+3y))$
answered Mar 8 '14 at 5:35
kmitovkmitov
4,0002816
$endgroup$
$begingroup$
I don't see any $5y$ in the question --- was there an edit?
$endgroup$
– Gerry Myerson
Mar 8 '14 at 7:20
$begingroup$
Yes, it is my mkistake. It is 3. I corrected it.
$endgroup$
– kmitov
Mar 8 '14 at 7:52
add a comment |
$begingroup$
$sin(4x+3y)+xcos(4x+3y)(4+3dy/dx)=(dy/dx)cos x -ysin x$
$sin(4x+3y)+4xcos(4x+3y)+3(dy/dx)cos(4x+3y)=(dy/dx)cos x -ysin x$
$sin(4x+3y)+4xcos(4x+3y)+ysin x=(dy/dx)cos x + 3(dy/dx)cos(4x+3y)$
$sin(4x+3y)+4xcos(4x+3y)+ysin x=(dy/dx)(cos x + 3cos(4x+3y))$
answered Mar 8 '14 at 5:35
kmitovkmitov
4,0002816
$endgroup$
$sin(4x+3y)+xcos(4x+3y)(4+3dy/dx)=(dy/dx)cos x -ysin x$
$sin(4x+3y)+4xcos(4x+3y)+3(dy/dx)cos(4x+3y)=(dy/dx)cos x -ysin x$
$sin(4x+3y)+4xcos(4x+3y)+ysin x=(dy/dx)cos x + 3(dy/dx)cos(4x+3y)$
$sin(4x+3y)+4xcos(4x+3y)+ysin x=(dy/dx)(cos x + 3cos(4x+3y))$
answered Mar 8 '14 at 5:35
kmitovkmitov
4,0002816
edited Mar 8 '14 at 7:52
answered Mar 8 '14 at 5:35
kmitovkmitov
4,0002816
answered Mar 8 '14 at 5:35
kmitovkmitov
4,0002816
answered Mar 8 '14 at 5:35
kmitovkmitov
4,0002816
4,0002816
$begingroup$
I don't see any $5y$ in the question --- was there an edit?
$endgroup$
– Gerry Myerson
Mar 8 '14 at 7:20
$begingroup$
Yes, it is my mkistake. It is 3. I corrected it.
$endgroup$
– kmitov
Mar 8 '14 at 7:52
add a comment |
$begingroup$
I don't see any $5y$ in the question --- was there an edit?
$endgroup$
– Gerry Myerson
Mar 8 '14 at 7:20
$begingroup$
Yes, it is my mkistake. It is 3. I corrected it.
$endgroup$
– kmitov
Mar 8 '14 at 7:52
$begingroup$
I don't see any $5y$ in the question --- was there an edit?
$endgroup$
– Gerry Myerson
Mar 8 '14 at 7:20
$begingroup$
I don't see any $5y$ in the question --- was there an edit?
$endgroup$
– Gerry Myerson
Mar 8 '14 at 7:20
$begingroup$
Yes, it is my mkistake. It is 3. I corrected it.
$endgroup$
– kmitov
Mar 8 '14 at 7:52
$begingroup$
Yes, it is my mkistake. It is 3. I corrected it.
$endgroup$
– kmitov
Mar 8 '14 at 7:52
add a comment |
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$begingroup$
You'll have to use the product rule on each side.
$endgroup$
– user61527
Mar 8 '14 at 5:27
$begingroup$
Write $F(x,y)$ as ($lhs-rhs$). Compute the partial derivatives of $F(x,y)$ with respect to $x$ and then to $y$ and compute the classical ratio from total differentiation.
$endgroup$
– Claude Leibovici
Mar 8 '14 at 5:33