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

The answer to (\partial)/(\partial x)((x^2-2y)/(4x^2+2y)) is (5xy)/((2x^2+y)^2)

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

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

\frac{5 x y}{( 2 x ^{2} + y ) ^{2}}

Solution steps

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

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

\frac{\partial}{\partial x} (4 x^{2} + 2 y) = 8 x

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

Simplify

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

= \frac{5 x y}{( 2 x ^{2} + y ) ^{2}}

\frac{\partial}{\partial y} (x y^{4})

f (x) = x^{3} - 3 x + 2, at x = 4

y = \frac{1}{x}, y = 2 x - 1, 0, 3

x \to \infty lim (4 cos (3 x))

\int 0.55 t^{2} + 1.575 t^{4} d t

What is (\partial)/(\partial x)((x^2-2y)/(4x^2+2y)) ?
The answer to (\partial)/(\partial x)((x^2-2y)/(4x^2+2y)) is (5xy)/((2x^2+y)^2)