Sorry I don't know how to do this either. I guess no one knows.
Hey guys here's the problem...
Suppose f is a differentiable function of x and y, and g(u,v) = f(e^u + sinv, e^u + cosv). Use the table of values to calculate gu(0,0) and gv(0,0).
At (0,0): f = 3, g = 6, fx = 4, fy = 8
At (1,2): f = 6, g = 3, fx = 2, fy = 5
Where fx and fy and f sub x and f sub y (partial derivatives).
My problem is, I can't figure out how to differentiate g(u,v) with respect to x or y. Any insight?
I want to be sure I understand which variables you are refering to and "what" you are refering to with certain terms:
I assume gu is the partial derivative of g with respect to u and gv is the same with respect to v.
I assume that the initial values (0,0) and (1,2) refer to the initial values of u and v, such that:
(u0,v0) = (0,0): f(u0,v0) = 3, g(u0,v0) = 6, fx(u0,v0) = 4, fy(u0,v0) = 8
(u1,v1) = (1,2): f(u1,v1) = 6, g(u1,v1) = 3, fx(u1,v1) = 2, fy(u1,v1) = 5
If my assumptions are correct, then I can do the problem:
Since f(x,y) is a function of x and y,
Where g(u,v) = f(e^u + sinv, e^u + cosv),
It holds that
x = e^u + sinv
y = e^u + cosv
Taking partial derivatives in terms of u, we get:
Du[g(u,v)] = Du[f(e^u + sinv, e^u + cosv)] = df/dx*dx/du + df/dy*dy/du
Notice that dx/du = e^u and dy/du = e^u, so
gu(u,v) = e^u*fx(u,v) + e^u*fy(u,v)
gu(0,0) = e^0*4 + e^0*8 = 4 + 8 = 12
Taking partial derivatives in terms of v, we get:
Dv[g(u,v)] = Dv[f(e^u + sinv, e^u + cosv)] = df/dx*dx/dv + df/dy*dy/dv
Notice that dx/dv = cosv and dy/dv = -sinv, so
gv(u,v) = cosv*fx(u,v) - sinv*fy(u,v)
gv(0,0) = cos0*4 - sin0*8 = 4
Edit: VERY BIG NOTE: the "d"s should be read as partial derivatives as all functions are in terms of more than one variable.
Who is the ImPerfectHacker? If this is ThePerfectHacker or CaptainBlack then I doubt there is much reason to care that this handle was used, but if not then the use of this handle is a clear violation of forum rules.
Edit: Upon reviewing the forum rules, it seems mocking a forum moderator's handle is not a "clear violation of forum rules" but it should be.
I think I should be more strict with the policy to post only math related stuff. No chat conversations. The Talk Spam Section is there for that.
But this was something else, a lot of begging and another evil move.
ImPerfectHacker is my good twin brother No seriously, it was a member having trying to get my attention. (Original Poster).Who is the ImPerfectHacker? If this is ThePerfectHacker or CaptainBlack then I doubt there is much reason to care that this handle was used, but if not then the use of this handle is a clear violation of forum rules.
Edit: Upon reviewing the forum rules, it seems mocking a forum moderator's handle is not a "clear violation of forum rules" but it should be.