What is (\partial)/(\partial z)(e^y+xcos(xz)) ?

The answer to (\partial)/(\partial z)(e^y+xcos(xz)) is -x^2sin(xz)

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(\partial)/(\partial z)(e^y+xcos(xz))

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

- x^{2} sin (x z)

Solution steps

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

Treat

y, x

as constants

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial z} (e^{y}) = 0

\frac{\partial}{\partial z} (x cos (x z)) = - x^{2} sin (x z)

= 0 - x^{2} sin (x z)

Simplify

= - x^{2} sin (x z)

f (x) = (x - 5 x x)

y^{^{'}} = 2 x + y, y (0) = 6

y = x^{2} + 1, y = x + 3

(x + 2 y) y^{^{'}} = 5 x - y

\int \frac{sin ( t )}{t} d t

What is (\partial)/(\partial z)(e^y+xcos(xz)) ?
The answer to (\partial)/(\partial z)(e^y+xcos(xz)) is -x^2sin(xz)