Is u(t) a unit vector of f(x,y)?

[Philip Wong]

Homework Statement

f(x,y)=2Sin x Cos y
g(x,y) = 2Cos x Sin y
verify that d(fg)/dx = g(x,y) df/dx + f(x,y) dg/dx

The Attempt at a Solution

first of all I worked out the partials derivatives in respective to x and y, for both functions
df/dx = 2Cos x (but I've a gut feeling that it should be 2Sin x * Cos x)
df/dy = -Sin y (but I've a gut feeling that it should be Cos y * -Sin y)
dg/dx=-2Sin x (but I've a gut feeling that it should be 2Cos x * -Sin x)
dg/dy=Cos y (but I've a gut feeling that it should be Sin y * Cos y)

but I don't get how to verify d(fg)/dx = g(x,y) df/dx + f(x,y) dg/dx
can someone walk me through what does the question is asking about?

Thanks

 

[Quinzio]

Just do what the problem asks you.

a) Multiply f and g toghether as they are, then derive for x.

b) Then compute g multiplied by df/dx plus f multiplied by dg/dx

Check if a) and b) are the same.

 

[HallsofIvy]

Homework Statement

f(x,y)=2Sin x Cos y
g(x,y) = 2Cos x Sin y
verify that d(fg)/dx = g(x,y) df/dx + f(x,y) dg/dx

The Attempt at a Solution

first of all I worked out the partials derivatives in respective to x and y, for both functions
df/dx = 2Cos x (but I've a gut feeling that it should be 2Sin x * Cos x)

Neither of those is correct. The derivative of 2A Sin x, where A is a constant, is 2A Cos x. Since in taking a partial derivative, you treat the other variable as a constant, the partial derivative of 2 Sin x Cos y is 2 Cos x Cos y.

df/dy = -Sin y (but I've a gut feeling that it should be Cos y * -Sin y)
dg/dx=-2Sin x (but I've a gut feeling that it should be 2Cos x * -Sin x)
dg/dy=Cos y (but I've a gut feeling that it should be Sin y * Cos y)

Same applies to each of those. (But you don't need the derivatives with respect to y for this problem.)

but I don't get how to verify d(fg)/dx = g(x,y) df/dx + f(x,y) dg/dx
can someone walk me through what does the question is asking about?

Thanks[/QUOTE]
fg= 4 Sin x Cos x Sin y Cos y. The left side of your equation is the derivative of that with respect to x.

 

[phyzguy]

You're not doing the partial derivatives correctly. When I take the partial derivative with respect to x, I treat y like a constant. So you should treat the cos(y) term just like you treat the 2. So the partial of {2 sin(x) cos(y)) with respect to x is 2 cos(x) cos(y), not 2 cos(x).

 

[Philip Wong]

You're not doing the partial derivatives correctly. When I take the partial derivative with respect to x, I treat y like a constant. So you should treat the cos(y) term just like you treat the 2. So the partial of {2 sin(x) cos(y)) with respect to x is 2 cos(x) cos(y), not 2 cos(x).

Neither of those is correct. The derivative of 2A Sin x, where A is a constant, is 2A Cos x. Since in taking a partial derivative, you treat the other variable as a constant, the partial derivative of 2 Sin x Cos y is 2 Cos x Cos y.


Same applies to each of those. (But you don't need the derivatives with respect to y for this problem.)

but I don't get how to verify d(fg)/dx = g(x,y) df/dx + f(x,y) dg/dx
can someone walk me through what does the question is asking about?

Thanks

fg= 4 Sin x Cos x Sin y Cos y. The left side of your equation is the derivative of that with respect to x.[/QUOTE]

oh right thank you very much! I think I need more practice on derivatives though.
another question, say I have the following function
f(x,y)=x²+y²-xy
I need to show that u(t)=(cos t; sin t) is a unit vector

can someone please tell me what steps I need to take to show u(t) is a unit vector of that given function?

thanks again