Find the first partial derivatives of the function. f(x,y)=ax+by/cx+dy

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Brighton · Accepted Answer

As a differentiation tool, use the quotient rule (while treating y like a constant)
$\frac{\partial f}{\partial x} = \frac{[\partial \frac{a x + b y}{\partial} x] \cdot (c x + d y) \cdot [\partial \frac{c x + d y}{\partial} x]}{{(c x + d y)}^{2}}$
$\frac{\partial f}{\partial x} = \frac{(a + 0) \cdot (c x + d y) - (a x + b y) \cdot (c + 0)}{{(c x + d y)}^{2}}$
$= \frac{a c x + a d y - c a x - c b y}{{(c x + d y)}^{2}} = \frac{(a d - b c) y}{{(c x + d y)}^{2}}$
As a differentiation tool, use the quotient rule (while treating x like a constant)
$\frac{\partial f}{\partial y} = \frac{[\partial \frac{a x + b y}{\partial} y] \cdot (c x + d y) \cdot [\partial \frac{c x + d y}{\partial} y]}{{(c x + d y)}^{2}}$
$\frac{\partial f}{\partial y} = \frac{(a + 0) \cdot (c x + d y) - (a x + b y) \cdot (0 + d)}{{(c x + d y)}^{2}}$
$= \frac{b c x + b d y - d a x - d b y}{{(c x + d y)}^{2}} = \frac{(b c - a d) x}{{(c x + d y)}^{2}}$
Result:
$f_{x} = \frac{(a d - b c) y}{{(c x + d y)}^{2}}$
$f_{y} = \frac{(b c - a d) x}{{(c x + d y)}^{2}}$

Find the first partial derivatives of the function. f(x,y)=ax+by/cx+dy

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