What is (\partial)/(\partial y)(((ax+by))/(cx+dy)) ?

The answer to (\partial)/(\partial y)(((ax+by))/(cx+dy)) is (xbc-axd)/((xc+yd)^2)

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(\partial)/(\partial y)(((ax+by))/(cx+dy))

\frac{\partial}{\partial y} (\frac{( a x + b y )}{c x + d y})

\frac{x b c - a x d}{( x c + y d ) ^{2}}

Solution steps

\frac{\partial}{\partial y} (\frac{( a x + b y )}{c x + d y})

Treat

a, x, b, c, d

as constants

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{\partial}{\partial y} ( a x + b y ) ( c x + d y ) - \frac{\partial}{\partial y} ( c x + d y ) ( a x + b y )}{( c x + d y ) ^{2}}

\frac{\partial}{\partial y} (a x + b y) = b

\frac{\partial}{\partial y} (c x + d y) = d

= \frac{b ( c x + d y ) - d ( a x + b y )}{( c x + d y ) ^{2}}

Expand

b (c x + d y) - d (a x + b y) : x b c - a x d

= \frac{x b c - a x d}{( x c + y d ) ^{2}}

What is (\partial)/(\partial y)(((ax+by))/(cx+dy)) ?
The answer to (\partial)/(\partial y)(((ax+by))/(cx+dy)) is (xbc-axd)/((xc+yd)^2)