What is (\partial)/(\partial y)((x)(2x+ycos(xy))) ?

The answer to (\partial)/(\partial y)((x)(2x+ycos(xy))) is x(cos(xy)-xysin(xy))

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(\partial)/(\partial y)((x)(2x+ycos(xy)))

\frac{\partial}{\partial y} ((x) (2 x + y cos (x y)))

x (cos (x y) - x y sin (x y))

Solution steps

\frac{\partial}{\partial y} ((x) (2 x + y cos (x y)))

Treat

x

as a constant

Take the constant out:

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

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial y} (2 x) = 0

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

= x (0 + cos (x y) - x y sin (x y))

Simplify

= x (cos (x y) - x y sin (x y))

f (p - \frac{1}{4})

\frac{( 2 h + 8 )}{( 7 h - 4 ) ( h - 1 )}

\frac{( - 2 + 6 ) 2 6 + 5}{2}

\frac{x + 10}{5 ( x + 2 )}

(x y^{2} - k x^{2} y) d x + (x - y) x^{2} d y

What is (\partial)/(\partial y)((x)(2x+ycos(xy))) ?
The answer to (\partial)/(\partial y)((x)(2x+ycos(xy))) is x(cos(xy)-xysin(xy))