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

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

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

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

- 2 sin (2 x + y)

Solution steps

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

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (z) = 0

= - sin (2 x + y) \cdot 2 - 0

Simplify

= - 2 sin (2 x + y)

\frac{d y}{d x} = \frac{6 y}{x}

x \to 0 lim (\frac{sin ( 6 x )}{2})

\int \frac{x}{3 x ^{2} - 5} d x

\frac{d y}{d x} = 3 x y^{2}

(x - y^{4} + y^{2} sin (x)) d x = (4 x y^{3} + 2 y cos (x)) d y

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