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

The answer to (\partial)/(\partial y)(2cos(x)+cos(x+y)) is -sin(x+y)

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(\partial)/(\partial y)(2cos(x)+cos(x+y))

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

- sin (x + y)

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 - sin (x + y)

Simplify

= - sin (x + y)

\int 5 sin^{2} (r) cos (r) d r

x^{2 x + 1}

x \to 5 lim (\frac{1}{x - 5})

\frac{d}{d x} ((ln (x))^{x})

(3, 12), (6, 13)

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