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

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

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

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

2 y - y sin (x)

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 2 y - y sin (x)

e^{8 x} - e^{2 x}, x = 1

y = x - 6, y^{2} = 2 x + 108

x \to 3 lim (\frac{13 sin ( 6 x )}{x})

\frac{d}{d x} (2 x - 15)

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