What is (\partial)/(\partial y)(y^2-2ycos(x)) ?

The answer to (\partial)/(\partial y)(y^2-2ycos(x)) is 2y-2cos(x)

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(\partial)/(\partial y)(y^2-2ycos(x))

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

2 y - 2 cos (x)

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 2 y - 2 cos (x)

ln (x e^{7 x})

f (x) = \frac{32}{x}

\frac{\partial}{\partial x} (4 π^{2} \frac{x}{y ^{2}})

x^{2} - 4 x + 3, 2 x^{2} + 2 x + 3, - 6, 0

\int \frac{1}{1 - 25 x ^{2}} d x

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