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+1 vote
114k views
in Differentials and Partial Derivatives by (49.1k points)
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If u = log(x2 + y2)/xy then x(∂u/∂x) + y(∂u/∂y) is ...

(a) 0 

(b) u 

(c) 2u 

(d) u – 1

2 Answers

+1 vote
by (17.0k points)
selected by
 
Best answer

Correct option is (a) 0

\(u =\log\left(\frac{x^2 + y^2}{xy}\right)\)

\(e^u = e^{\log}\left(\frac{x^2 + y^2}{xy}\right)\)

\(f\) ⇒ \(e^u = \left(\frac{x^2 + y^2}{xy}\right)\)

\(f(x,y) = \frac{x^2 +y^2}{x y}\)

\(f(xt,yt) = \frac{(xt)^2 +(yt)^2}{x tyt}\)

\(= \frac{t^2(x^2 + y^2)}{t^2 (xy)}\)

\(f(xt, yt) = t^{2-2}.f(x, y)\)

\(f(xt, yt) = t^0.f(x, y)\)

This is the form of f(xt, yt) = tn f(x, y)

n = 0

f is a homogeneous function of degree n = 0

Euler's theorem:

\(x\frac {\partial f}{\partial x} + y\frac{\partial f}{\partial y} = n f \)

\(x\frac {\partial }{\partial x} (e^u)+ y\frac{\partial }{\partial y} (e^u)= 0\times f \)

\(x \,e^u \frac{\partial u}{\partial x} + y\,e^u \frac{\partial u}{\partial y} = 0\)

\(e^u \left(x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}\right) =0\)

\(x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} =0\)

+3 votes
by (46.9k points)

Answer is (a) 0

Given 

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