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

The answer to (\partial)/(\partial x)((4x-2y)/(2x^2+3y)) is (-8x^2+8xy+12y)/((2x^2+3y)^2)

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(\partial)/(\partial x)((4x-2y)/(2x^2+3y))

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

\frac{- 8 x ^{2} + 8 x y + 12 y}{( 2 x ^{2} + 3 y ) ^{2}}

Solution steps

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

\frac{\partial}{\partial x} (4 x - 2 y) = 4

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

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

Expand

4 (2 x^{2} + 3 y) - 4 x (4 x - 2 y) : - 8 x^{2} + 8 x y + 12 y

= \frac{- 8 x ^{2} + 8 x y + 12 y}{( 2 x ^{2} + 3 y ) ^{2}}

\int_{0}^{4} \frac{1}{x ^{2} + x - 6} d x

\int_{1}^{4} (4 x^{2} + x) d x

\int \frac{1}{1 - 5 t} d t

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

\int_{0}^{2} ∣ x - 1 ∣ d x

What is (\partial)/(\partial x)((4x-2y)/(2x^2+3y)) ?
The answer to (\partial)/(\partial x)((4x-2y)/(2x^2+3y)) is (-8x^2+8xy+12y)/((2x^2+3y)^2)