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

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

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

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

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

Solution steps

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

\frac{\partial}{\partial x} (x) = 1

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

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

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

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

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

\frac{\partial}{\partial x} (y cos (x - y))

64 y^{^{''}} + 80 y^{^{'}} + 25 y = 0, y (0) = a, y^{^{'}} (0) = - 5

f (x) = ln (x^{2} - 9 x + 1), at x = 9

\int e^{- x - y} d x

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