What is (\partial)/(\partial y)(cos(x)+xcos(y)-y) ?

The answer to (\partial)/(\partial y)(cos(x)+xcos(y)-y) is -xsin(y)-1

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(\partial)/(\partial y)(cos(x)+xcos(y)-y)

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

- x sin (y) - 1

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 0 - x sin (y) - 1

Simplify

= - x sin (y) - 1

n = 0 \sum \infty (\frac{2}{4})^{n}

\int 5 x + 12 d x

n = 0 \sum \infty n \cdot (\frac{1}{6})^{n}

x y^{^{'}} + 2 y - 2 x^{2} y^{2} = 0

f (x) = (2 x - 3)^{4} (x^{2} + x + 1)^{5}

What is (\partial)/(\partial y)(cos(x)+xcos(y)-y) ?
The answer to (\partial)/(\partial y)(cos(x)+xcos(y)-y) is -xsin(y)-1