What is (\partial)/(\partial x)((31xy)/(x-y)) ?

The answer to (\partial)/(\partial x)((31xy)/(x-y)) is -(31y^2)/((x-y)^2)

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(\partial)/(\partial x)((31xy)/(x-y))

\frac{\partial}{\partial x} (\frac{31 x y}{x - y})

- \frac{31 y ^{2}}{( x - y ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{31 x y}{x - y})

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 31 y \frac{\partial}{\partial x} (\frac{x}{x - y})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= 31 y \frac{\frac{\partial}{\partial x} ( x ) ( x - y ) - \frac{\partial}{\partial x} ( x - y ) x}{( x - y ) ^{2}}

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

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

= 31 y \frac{1 \cdot ( x - y ) - 1 \cdot x}{( x - y ) ^{2}}

Simplify

31 y \frac{1 \cdot ( x - y ) - 1 \cdot x}{( x - y ) ^{2}} : - \frac{31 y ^{2}}{( x - y ) ^{2}}

= - \frac{31 y ^{2}}{( x - y ) ^{2}}

\int sec^{2} (θ) tan^{8} (θ) d θ

\frac{\partial}{\partial x} (1 + y^{2})

9 y^{^{''}} - 6 y^{^{'}} + y = 0

n = 1 \sum \infty (\frac{1}{7})^{n}

4 x - x^{2}, x^{2}, 2, 0

What is (\partial)/(\partial x)((31xy)/(x-y)) ?
The answer to (\partial)/(\partial x)((31xy)/(x-y)) is -(31y^2)/((x-y)^2)