Z = xyln(xy) ∂ z∂ x=xy ( yxy )+yln(xy) = y(1 + lnxy) Similarly, nbsp; nbsp; ∂ z∂ y=x(1+lnxy) x∂ z∂ x=xy(1+lnxy) y∂ z∂ y=xy(1+lnxy) So x∂ z∂ x= y∂ z ...

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Standard X

Mathematics

Method of Substitution to Find the Solution of a Pair of Linear Equations

If z = xy Inx...

Question

If z = xy In(xy), then

x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 0

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y \frac{\partial z}{\partial x} = x \frac{\partial z}{\partial y}

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x \frac{\partial z}{\partial x} = y \frac{\partial z}{\partial y}

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y \frac{\partial z}{\partial x} + x \frac{\partial z}{\partial y} = 0

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Solution

The correct option is C $x \frac{\partial z}{\partial x} = y \frac{\partial z}{\partial y}$
z = xyln(xy)

$\frac{\partial z}{\partial x} = x y (\frac{y}{x y}) + y l n (x y)$

$= y (1 + l n x y) S i m i l a r l y, \frac{\partial z}{\partial y} = x (1 + l n x y)$

$x \frac{\partial z}{\partial x} = x y (1 + l n x y)$

$y \frac{\partial z}{\partial y} = x y (1 + l n x y)$

So $x \frac{\partial z}{\partial x} = y \frac{\partial z}{\partial y}$

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If z = xy In(xy), then

The correct option is C x∂z∂x=y∂z∂yz = xyln(xy) ∂z∂x=xy(yxy)+yln(xy) =y(1+lnxy)Similarly,∂z∂y=x(1+lnxy) x∂z∂x=xy(1+lnxy) y∂z∂y=xy(1+lnxy) So x∂z∂x=y∂z∂y