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

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

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

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

yz + sin (x + y + z)

Solution steps

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

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (x yz) = yz

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

= yz - (- sin (x + y + z))

Simplify

= yz + sin (x + y + z)

\int - g e^{- c t} \frac{1}{c} d t

\frac{\partial}{\partial x} (4 x^{2} + 2 x y - 8 y^{2})

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

\frac{\partial}{\partial x} (tan (4 + 2 x^{2} y^{4} z^{2}))

\int (- 3 cos (x) + 4 sec^{2} (x)) d x

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