What is (\partial)/(\partial y)(-(sin(2xy))/(xy)) ?

The answer to (\partial)/(\partial y)(-(sin(2xy))/(xy)) is -(2xycos(2xy)-sin(2xy))/(xy^2)

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(\partial)/(\partial y)(-(sin(2xy))/(xy))

\frac{\partial}{\partial y} (- \frac{sin ( 2 x y )}{x y})

- \frac{2 x y cos ( 2 x y ) - sin ( 2 x y )}{x y ^{2}}

Solution steps

\frac{\partial}{\partial y} (- \frac{sin ( 2 x y )}{x y})

Treat

x

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{\partial}{\partial y} (sin (2 x y)) = cos (2 x y) \cdot 2 x

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

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

Simplify

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

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

x^{- \frac{3}{4}} + x^{\frac{3}{4}}

\frac{d}{d x} (2 ln (2 x + 5))

\frac{d}{d x} (\frac{x}{1 + 2 x})

\int \frac{x}{x ^{4} + 25} d x

\int \frac{5}{x} d x

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