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

The answer to (\partial)/(\partial x)((y-5x)/(y^2+4x^2)) is (20x^2-8yx-5y^2)/((y^2+4x^2)^2)

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(\partial)/(\partial x)((y-5x)/(y^2+4x^2))

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

\frac{20 x ^{2} - 8 y x - 5 y ^{2}}{( y ^{2} + 4 x ^{2} ) ^{2}}

Solution steps

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

\frac{\partial}{\partial x} (y - 5 x) = - 5

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

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

Simplify

\frac{( - 5 ) ( y ^{2} + 4 x ^{2} ) - 8 x ( y - 5 x )}{( y ^{2} + 4 x ^{2} ) ^{2}} : \frac{20 x ^{2} - 8 y x - 5 y ^{2}}{( y ^{2} + 4 x ^{2} ) ^{2}}

= \frac{20 x ^{2} - 8 y x - 5 y ^{2}}{( y ^{2} + 4 x ^{2} ) ^{2}}

\frac{\partial}{\partial x} (x^{0.5} y)

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = 4

\int_{0}^{3} e^{x} d x

f (x) = x + 16, at x = 1

\frac{d y}{d x} = (y)

What is (\partial)/(\partial x)((y-5x)/(y^2+4x^2)) ?
The answer to (\partial)/(\partial x)((y-5x)/(y^2+4x^2)) is (20x^2-8yx-5y^2)/((y^2+4x^2)^2)