What is (\partial)/(\partial y)(4e^xcos(yz)) ?

The answer to (\partial)/(\partial y)(4e^xcos(yz)) is -4e^xzsin(yz)

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(\partial)/(\partial y)(4e^xcos(yz))

\frac{\partial}{\partial y} (4 e^{x} cos (yz))

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

Solution steps

\frac{\partial}{\partial y} (4 e^{x} cos (yz))

Treat

x, z

as constants

Take the constant out:

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

= 4 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

= 4 e^{x} (- sin (yz) z)

Simplify

= - 4 e^{x} z sin (yz)

x \to \infty lim \frac{( ( - 4 ) ^{x + 1} )}{( x + 1 ) \cdot 4 ^{x + 1}}

\frac{d}{d x} (4 sec (\frac{x}{2}))

\int sec^{2} (x) - 8 cot (x) d x

y^{^{''}} + 36 y = 24 sin (6 t)

y = \frac{1}{7 t ^{3}}

What is (\partial)/(\partial y)(4e^xcos(yz)) ?
The answer to (\partial)/(\partial y)(4e^xcos(yz)) is -4e^xzsin(yz)