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

The answer to (\partial)/(\partial x)((-3x-y)/(y^2+2x^2)) is (-3y^2+6x^2+4xy)/((y^2+2x^2)^2)

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

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

\frac{- 3 y ^{2} + 6 x ^{2} + 4 x y}{( y ^{2} + 2 x ^{2} ) ^{2}}

Solution steps

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

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

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

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

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

Simplify

\frac{( - 3 ) ( y ^{2} + 2 x ^{2} ) - 4 x ( - 3 x - y )}{( y ^{2} + 2 x ^{2} ) ^{2}} : \frac{- 3 y ^{2} + 6 x ^{2} + 4 x y}{( y ^{2} + 2 x ^{2} ) ^{2}}

= \frac{- 3 y ^{2} + 6 x ^{2} + 4 x y}{( y ^{2} + 2 x ^{2} ) ^{2}}

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

y^{^{'}} = 20 x - 2

\frac{d}{d x} (2 x + e^{x} + e^{- x})

F (x) = sin (7 x) + cos (4 x)

\frac{d}{d x} (\frac{10 x}{x ^{2} + 1})

What is (\partial)/(\partial x)((-3x-y)/(y^2+2x^2)) ?
The answer to (\partial)/(\partial x)((-3x-y)/(y^2+2x^2)) is (-3y^2+6x^2+4xy)/((y^2+2x^2)^2)