Mixing
Differentiate with respect to $y$: $f(x,y,z) = sin(x)sin(y)sin(z)-cos(x)cos(y)cos(z)$
$begingroup$
The derivative with respect to $y$ of
$$f(x,y,z) = sin(x)sin(y)sin(z)-cos(x)cos(y)cos(z)$$
is
a) $f’(x,y,z) = cos(x)cos(y)sin(z) + sin(x)sin(y)cos(z)$
b) $f’(x,y,z) = sin(x)cos(y)sin(z) + cos(x)sin(y)cos(z)$
c) $f’(x,y,z) = cos(x)cos(y)cos(z) + sin(x)sin(y)sin(z)$
d) $f’(x,y,z) = sin(x)sin(y)sin(z) + cos(x)cos(y)cos(z)$
trigonometry
edited Jan 15 at 13:53
Blue
48.5k870154
asked Jan 15 at 11:39
JohnJohn
1
$endgroup$
add a comment |
$begingroup$
The derivative with respect to $y$ of
$$f(x,y,z) = sin(x)sin(y)sin(z)-cos(x)cos(y)cos(z)$$
is
a) $f’(x,y,z) = cos(x)cos(y)sin(z) + sin(x)sin(y)cos(z)$
b) $f’(x,y,z) = sin(x)cos(y)sin(z) + cos(x)sin(y)cos(z)$
c) $f’(x,y,z) = cos(x)cos(y)cos(z) + sin(x)sin(y)sin(z)$
d) $f’(x,y,z) = sin(x)sin(y)sin(z) + cos(x)cos(y)cos(z)$
trigonometry
edited Jan 15 at 13:53
Blue
48.5k870154
asked Jan 15 at 11:39
JohnJohn
1
$endgroup$
1
$begingroup$
What are your thoughts? What have you tried? Where did you get stuck? Questions that just state a problem are likely to be downvoted and closed.
$endgroup$
– postmortes
Jan 15 at 12:04
add a comment |
$begingroup$
The derivative with respect to $y$ of
$$f(x,y,z) = sin(x)sin(y)sin(z)-cos(x)cos(y)cos(z)$$
is
a) $f’(x,y,z) = cos(x)cos(y)sin(z) + sin(x)sin(y)cos(z)$
b) $f’(x,y,z) = sin(x)cos(y)sin(z) + cos(x)sin(y)cos(z)$
c) $f’(x,y,z) = cos(x)cos(y)cos(z) + sin(x)sin(y)sin(z)$
d) $f’(x,y,z) = sin(x)sin(y)sin(z) + cos(x)cos(y)cos(z)$
trigonometry
edited Jan 15 at 13:53
Blue
48.5k870154
asked Jan 15 at 11:39
JohnJohn
1
$endgroup$
The derivative with respect to $y$ of
$$f(x,y,z) = sin(x)sin(y)sin(z)-cos(x)cos(y)cos(z)$$
is
a) $f’(x,y,z) = cos(x)cos(y)sin(z) + sin(x)sin(y)cos(z)$
b) $f’(x,y,z) = sin(x)cos(y)sin(z) + cos(x)sin(y)cos(z)$
c) $f’(x,y,z) = cos(x)cos(y)cos(z) + sin(x)sin(y)sin(z)$
d) $f’(x,y,z) = sin(x)sin(y)sin(z) + cos(x)cos(y)cos(z)$
trigonometry
trigonometry
edited Jan 15 at 13:53
Blue
48.5k870154
asked Jan 15 at 11:39
JohnJohn
1
edited Jan 15 at 13:53
Blue
48.5k870154
asked Jan 15 at 11:39
JohnJohn
1
edited Jan 15 at 13:53
Blue
48.5k870154
edited Jan 15 at 13:53
Blue
48.5k870154
edited Jan 15 at 13:53
Blue
48.5k870154
48.5k870154
asked Jan 15 at 11:39
JohnJohn
1
asked Jan 15 at 11:39
JohnJohn
1
asked Jan 15 at 11:39
JohnJohn
1
1
1
$begingroup$
What are your thoughts? What have you tried? Where did you get stuck? Questions that just state a problem are likely to be downvoted and closed.
$endgroup$
– postmortes
Jan 15 at 12:04
add a comment |
1
$begingroup$
What are your thoughts? What have you tried? Where did you get stuck? Questions that just state a problem are likely to be downvoted and closed.
$endgroup$
– postmortes
Jan 15 at 12:04
1
1
$begingroup$
What are your thoughts? What have you tried? Where did you get stuck? Questions that just state a problem are likely to be downvoted and closed.
$endgroup$
– postmortes
Jan 15 at 12:04
$begingroup$
What are your thoughts? What have you tried? Where did you get stuck? Questions that just state a problem are likely to be downvoted and closed.
$endgroup$
– postmortes
Jan 15 at 12:04
add a comment |
1 Answer
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$begingroup$
If we differentiate with respect to $y$ the other variables are assumed to be constant, so we get
$$frac{partial f(x,y,z)}{partial y}=cos(y)sin(x)sin(z)+sin(y)cos(x)cos(z)$$
answered Jan 15 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.7k42866
$endgroup$
add a comment |
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1 Answer
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$begingroup$
If we differentiate with respect to $y$ the other variables are assumed to be constant, so we get
$$frac{partial f(x,y,z)}{partial y}=cos(y)sin(x)sin(z)+sin(y)cos(x)cos(z)$$
answered Jan 15 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.7k42866
$endgroup$
add a comment |
$begingroup$
If we differentiate with respect to $y$ the other variables are assumed to be constant, so we get
$$frac{partial f(x,y,z)}{partial y}=cos(y)sin(x)sin(z)+sin(y)cos(x)cos(z)$$
answered Jan 15 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.7k42866
$endgroup$
add a comment |
$begingroup$
If we differentiate with respect to $y$ the other variables are assumed to be constant, so we get
$$frac{partial f(x,y,z)}{partial y}=cos(y)sin(x)sin(z)+sin(y)cos(x)cos(z)$$
answered Jan 15 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.7k42866
$endgroup$
If we differentiate with respect to $y$ the other variables are assumed to be constant, so we get
$$frac{partial f(x,y,z)}{partial y}=cos(y)sin(x)sin(z)+sin(y)cos(x)cos(z)$$
answered Jan 15 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.7k42866
answered Jan 15 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.7k42866
answered Jan 15 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.7k42866
answered Jan 15 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.7k42866
75.7k42866
add a comment |
add a comment |
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$begingroup$
What are your thoughts? What have you tried? Where did you get stuck? Questions that just state a problem are likely to be downvoted and closed.
$endgroup$
– postmortes
Jan 15 at 12:04