What is (\partial)/(\partial x)(-(8x-4y)/(-5x-9y)) ?

The answer to (\partial)/(\partial x)(-(8x-4y)/(-5x-9y)) is (92y)/((-5x-9y)^2)

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(\partial)/(\partial x)(-(8x-4y)/(-5x-9y))

\frac{\partial}{\partial x} (- \frac{8 x - 4 y}{- 5 x - 9 y})

\frac{92 y}{( - 5 x - 9 y ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (- \frac{8 x - 4 y}{- 5 x - 9 y})

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - \frac{\partial}{\partial x} (\frac{8 x - 4 y}{- 5 x - 9 y})

Apply the Quotient Rule:

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

= - \frac{\frac{\partial}{\partial x} ( 8 x - 4 y ) ( - 5 x - 9 y ) - \frac{\partial}{\partial x} ( - 5 x - 9 y ) ( 8 x - 4 y )}{( - 5 x - 9 y ) ^{2}}

\frac{\partial}{\partial x} (8 x - 4 y) = 8

\frac{\partial}{\partial x} (- 5 x - 9 y) = - 5

= - \frac{8 ( - 5 x - 9 y ) - ( - 5 ) ( 8 x - 4 y )}{( - 5 x - 9 y ) ^{2}}

Simplify

- \frac{8 ( - 5 x - 9 y ) - ( - 5 ) ( 8 x - 4 y )}{( - 5 x - 9 y ) ^{2}} : \frac{92 y}{( - 5 x - 9 y ) ^{2}}

= \frac{92 y}{( - 5 x - 9 y ) ^{2}}

\frac{d}{d x} (\frac{2 x ^{5} - 15}{x ^{4}})

e^{6 y}

\frac{d}{d x} (sin (3 x) + cos (2 x))

What is (\partial)/(\partial x)(-(8x-4y)/(-5x-9y)) ?
The answer to (\partial)/(\partial x)(-(8x-4y)/(-5x-9y)) is (92y)/((-5x-9y)^2)