Question

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Answer 1

$z = x \sqrt{y} + y \sqrt{x}$

$\frac{dz}{dx} = (x \sqrt{y} + y \sqrt{x} )'_{x} = \sqrt{y} + \frac{y}{2 \sqrt{x} } \\ \frac{ {d}^{2}z }{d {x}^{2} } = ( \sqrt{y } + \frac{y}{2 \sqrt{x} } )'_{x} = 0 -y \times \frac{2 \times \frac{1}{2 \sqrt{x} } }{( {2 \sqrt{x} })^{2}} = - \frac{y}{4x \sqrt{x} } \\ \frac{dz}{dy} = (x \sqrt{y} + y \sqrt{x} )'_{y} = \frac{x}{2 \sqrt{y} } + \sqrt{x} \\ \frac{ {d}^{2}z }{d {y}^{2} } = ( \frac{x}{2 \sqrt{y} } + \sqrt{x} )'_{y} = 0 - x \times \frac{2 \times \frac{1}{2 \sqrt{y} } }{( {2 \sqrt{y}) }^{2} } = - \frac{x }{4y \sqrt{y} } \\ \frac{ {d}^{2} z}{dxdy} = ( \sqrt{y} + \frac{y}{2 \sqrt{x} })'_{y} = \frac{1}{2 \sqrt{x} } + \frac{1}{2 \sqrt{y} } \\ \frac{ {d}^{2} z}{dydx} = ( \frac{x}{2 \sqrt{y} } + \sqrt{x} )'_{x} = \frac{1}{2 \sqrt{x} } + \frac{1}{2 \sqrt{y} }$