What is (\partial)/(\partial x)((6x-7y)/(y^2+x^2)) ?

The answer to (\partial)/(\partial x)((6x-7y)/(y^2+x^2)) is (6y^2-6x^2+14xy)/((y^2+x^2)^2)

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(\partial)/(\partial x)((6x-7y)/(y^2+x^2))

\frac{\partial}{\partial x} (\frac{6 x - 7 y}{y ^{2} + x ^{2}})

\frac{6 y ^{2} - 6 x ^{2} + 14 x y}{( y ^{2} + x ^{2} ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{6 x - 7 y}{y ^{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} ( 6 x - 7 y ) ( y ^{2} + x ^{2} ) - \frac{\partial}{\partial x} ( y ^{2} + x ^{2} ) ( 6 x - 7 y )}{( y ^{2} + x ^{2} ) ^{2}}

\frac{\partial}{\partial x} (6 x - 7 y) = 6

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

= \frac{6 ( y ^{2} + x ^{2} ) - 2 x ( 6 x - 7 y )}{( y ^{2} + x ^{2} ) ^{2}}

Expand

6 (y^{2} + x^{2}) - 2 x (6 x - 7 y) : 6 y^{2} - 6 x^{2} + 14 x y

= \frac{6 y ^{2} - 6 x ^{2} + 14 x y}{( y ^{2} + x ^{2} ) ^{2}}

x y^{^{'}} = y + x^{2} sin (x), y (\frac{π}{6}) = 0

\frac{d}{d x} ((\frac{x ^{2} + 2}{x ^{3} - 1})^{3})

\frac{1}{s ^{3} ( s ^{2} + 1 )}

\frac{d}{d x} (3 x^{2} ln (x) + x^{2})

\int - \frac{1}{sin ( y )} d y

What is (\partial)/(\partial x)((6x-7y)/(y^2+x^2)) ?
The answer to (\partial)/(\partial x)((6x-7y)/(y^2+x^2)) is (6y^2-6x^2+14xy)/((y^2+x^2)^2)