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

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

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

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

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

Solution steps

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

\frac{\partial}{\partial x} (x - 2 y) = 1

\frac{\partial}{\partial x} (y^{2} + 5 x^{2}) = 10 x

= \frac{1 \cdot ( y ^{2} + 5 x ^{2} ) - 10 x ( x - 2 y )}{( y ^{2} + 5 x ^{2} ) ^{2}}

Simplify

\frac{1 \cdot ( y ^{2} + 5 x ^{2} ) - 10 x ( x - 2 y )}{( y ^{2} + 5 x ^{2} ) ^{2}} : \frac{y ^{2} - 5 x ^{2} + 20 x y}{( y ^{2} + 5 x ^{2} ) ^{2}}

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

\frac{\partial}{\partial x} (7 x + 16)

\frac{1}{2} (6 t + 4)^{- \frac{1}{2}}

\int_{0}^{7} \frac{1}{( 6 - 2 x ) ^{\frac{1}{3}}} d x

\int 7 tan^{2} (x) sec^{3} (x) d x

\frac{d}{d x} (2 x^{3} - 6 x + 5)

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