What is (\partial)/(\partial x)((-6x+1y)/(-8x-1y)) ?

The answer to (\partial)/(\partial x)((-6x+1y)/(-8x-1y)) is (14y)/((-8x-y)^2)

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(\partial)/(\partial x)((-6x+1y)/(-8x-1y))

\frac{\partial}{\partial x} (\frac{- 6 x + 1 y}{- 8 x - 1 y})

\frac{14 y}{( - 8 x - y ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{- 6 x + 1 \cdot y}{- 8 x - 1 \cdot y})

Treat

y

as a constant

Apply the Quotient Rule:

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

= \frac{\frac{\partial}{\partial x} ( - 6 x + 1 \cdot y ) ( - 8 x - 1 \cdot y ) - \frac{\partial}{\partial x} ( - 8 x - 1 \cdot y ) ( - 6 x + 1 \cdot y )}{( - 8 x - 1 \cdot y ) ^{2}}

\frac{\partial}{\partial x} (- 6 x + 1 \cdot y) = - 6

\frac{\partial}{\partial x} (- 8 x - 1 \cdot y) = - 8

= \frac{( - 6 ) ( - 8 x - 1 \cdot y ) - ( - 8 ) ( - 6 x + 1 \cdot y )}{( - 8 x - 1 \cdot y ) ^{2}}

Simplify

\frac{( - 6 ) ( - 8 x - 1 \cdot y ) - ( - 8 ) ( - 6 x + 1 \cdot y )}{( - 8 x - 1 \cdot y ) ^{2}} : \frac{14 y}{( - 8 x - y ) ^{2}}

= \frac{14 y}{( - 8 x - y ) ^{2}}

\int_{0}^{1} \frac{2}{3 x ^{2} + 4 x + 1} d x

x \to 0 lim (\frac{6 ^{x} - 4 ^{x}}{x})

\int - x^{2} \frac{3}{3} d x

\frac{d}{d x} (x tan (y))

\int_{0}^{2} \frac{1}{x ^{2} + 4} d x

What is (\partial)/(\partial x)((-6x+1y)/(-8x-1y)) ?
The answer to (\partial)/(\partial x)((-6x+1y)/(-8x-1y)) is (14y)/((-8x-y)^2)