What is (\partial)/(\partial y)(9e^xcos(yz)) ?

The answer to (\partial)/(\partial y)(9e^xcos(yz)) is -9e^xzsin(yz)

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(\partial)/(\partial y)(9e^xcos(yz))

\frac{\partial}{\partial y} (9 e^{x} cos (yz))

- 9 e^{x} z sin (yz)

Solution steps

\frac{\partial}{\partial y} (9 e^{x} cos (yz))

Treat

x, z

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 9 e^{x} \frac{\partial}{\partial y} (cos (yz))

Apply the chain rule:

- sin (yz) \frac{\partial}{\partial y} (yz)

= - sin (yz) \frac{\partial}{\partial y} (yz)

\frac{\partial}{\partial y} (yz) = z

= 9 e^{x} (- sin (yz) z)

Simplify

= - 9 e^{x} z sin (yz)

(f (x) e^{g (x)})^{'}

\frac{d}{d x} (x - x)

\frac{d ^{3}}{d x ^{3}} (e^{2 x})

x \to 0 lim (\frac{1 - x}{x ^{2}})

y^{^{''}} + 4 y = 20

What is (\partial)/(\partial y)(9e^xcos(yz)) ?
The answer to (\partial)/(\partial y)(9e^xcos(yz)) is -9e^xzsin(yz)