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u = sin^{1}(x + y/(√x + √y))
Put x = xt and y = yt
u = sin^{}^{1}[(xtyt)/(√xt + √yt)]
sinu = t^{1/2}[(x+y)/(√x + √y)]
= t^{1/2} f(x,y)
This function sin u is homogeneous with degree 1/2.
By Euler's theorem
xu_{x} + yu_{y} = G(u) = nf(u)/f,(u) = 1/2 tan u
xu_{x} + yu_{y} = 1/2 tan u
x2u_{xx} + 2xyu_{yy} + y2u_{yy} = G(u)[G'(u)  1]
= 1/2 tan u [(sec^{2}u  2)/2]
= 1/4 tan u [(tan^{2}u  1)/1]
= 1/4 * sin u/cos u [(sin^{2}u  cos^{2}u)/cos^{2}u]
x^{2}u_{xx} + 2xyu_{yy} + y2u_{yy} = (sinu cos^{2}u)/4ucos^{3}u
then the value of is equal to
We have
v is homogeneous function of degree n then
The correct answer is: n(n  1)
The accompanying figure shows the graph of an unspecified function of f(x, y) and partial derivatives f_{x}(x, y) and f_{y}(x, y). Determine which is which and explain.
The correct answer is: II  f(x, y), I  f_{x}(x, y), III  f_{y}(x, y)
We are given
f(u) is homogeneous function degree 1, then
[By dividing with cos^{3} u]
The correct answer is:
If u = f(t) and v = f(t) and t = φ (x, y) then
The correct answer is:
If w = f(u, v) where u = x + y and v = x – y then
We have w = f(u, v)
The correct answer is:
If f is differentiable and z = u + f(u^{2}v^{2}), then
Let w = u^{2}v^{2}
then z = f(w) + u
So,
The correct answer is:
f(u) is homogeneous function of degree 2
Now, let g(u) = 2tan u
The correct answer is: 2tan u(2sec^{2}u –1)
If f(x, y, z) = 0 then the value of equal to
(1) Differentiate with respect to y, I get:
0+F2+F3∂z/∂y=0
So
F3 ∂z/∂y = −F2
(2) Differentiate with respect to x, I get:
F1 + F2 ∂y/∂x + 0 = 0
So F2 ∂y/∂x = −F1
(3) Differentiate with respect to z, I get:
F1 ∂x/∂z +0 + F3 = 0
4) After some manipulations with the Fi, I get to the conclusion that
∂z/∂y∗∂y/∂x∗∂x/∂z=−1, so when evaluated with x, z, y respectively, conclusion is still true
Use the information the figure to find the first order partial derivatives of f at the point (1, 2)
The correct answer is:
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