What is (\partial)/(\partial x)(((4x+y))/(x-3y)) ?

The answer to (\partial)/(\partial x)(((4x+y))/(x-3y)) is -(13y)/((x-3y)^2)

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(\partial)/(\partial x)(((4x+y))/(x-3y))

\frac{\partial}{\partial x} (\frac{( 4 x + y )}{x - 3 y})

- \frac{13 y}{( x - 3 y ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{( 4 x + y )}{x - 3 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} ( 4 x + y ) ( x - 3 y ) - \frac{\partial}{\partial x} ( x - 3 y ) ( 4 x + y )}{( x - 3 y ) ^{2}}

\frac{\partial}{\partial x} (4 x + y) = 4

\frac{\partial}{\partial x} (x - 3 y) = 1

= \frac{4 ( x - 3 y ) - 1 \cdot ( 4 x + y )}{( x - 3 y ) ^{2}}

Simplify

\frac{4 ( x - 3 y ) - 1 \cdot ( 4 x + y )}{( x - 3 y ) ^{2}} : - \frac{13 y}{( x - 3 y ) ^{2}}

= - \frac{13 y}{( x - 3 y ) ^{2}}

\frac{d}{d x} (arctan (\frac{x}{2}))

\int \frac{8 x ^{7} + 2 x}{x ^{8} + x ^{2}} d x

\frac{\partial}{\partial y} (\frac{2 x}{y ^{2}})

y^{^{'}} = 8 (y^{2} + 1)

\int x tan (x^{2} - 1) d x

What is (\partial)/(\partial x)(((4x+y))/(x-3y)) ?
The answer to (\partial)/(\partial x)(((4x+y))/(x-3y)) is -(13y)/((x-3y)^2)