What is (\partial)/(\partial x)(e^{xz+6}-cos(x+y)) ?

The answer to (\partial)/(\partial x)(e^{xz+6}-cos(x+y)) is e^{xz+6}z+sin(x+y)

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(\partial)/(\partial x)(e^{xz+6}-cos(x+y))

\frac{\partial}{\partial x} (e^{x z + 6} - cos (x + y))

e^{x z + 6} z + sin (x + y)

Solution steps

\frac{\partial}{\partial x} (e^{x z + 6} - cos (x + y))

Treat

z, y

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (e^{x z + 6}) - \frac{\partial}{\partial x} (cos (x + y))

\frac{\partial}{\partial x} (e^{x z + 6}) = e^{x z + 6} z

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

= e^{x z + 6} z - (- sin (x + y))

Simplify

= e^{x z + 6} z + sin (x + y)

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What is (\partial)/(\partial x)(e^{xz+6}-cos(x+y)) ?
The answer to (\partial)/(\partial x)(e^{xz+6}-cos(x+y)) is e^{xz+6}z+sin(x+y)