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

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

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

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

2 y sin (x)

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 - (- 2 y sin (x))

Simplify

= 2 y sin (x)

\frac{d}{d t} (v (t))

\frac{d}{d x} (\frac{f ( x )}{g ( x )})

y^{^{''}} + 9 y = 2 cos (3 x)

\frac{1}{4} e^{3 x}

x \to 1 - lim (x^{2} - 1)

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