What is (\partial)/(\partial y)(2e^xcos(yz)) ?

The answer to (\partial)/(\partial y)(2e^xcos(yz)) is -2e^xzsin(yz)

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(\partial)/(\partial y)(2e^xcos(yz))

\frac{\partial}{\partial y} (2 e^{x} cos (yz))

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

Solution steps

\frac{\partial}{\partial y} (2 e^{x} cos (yz))

Treat

x, z

as constants

Take the constant out:

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

= 2 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

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

Simplify

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

\frac{1}{3} (\frac{d x}{d t}) - 2 e^{- t} = - x

e^{2 x} tan (x)

(1 - x)^{- \frac{1}{2}}

\frac{d y}{d t} + y = \frac{1}{1 + e ^{2 t}}

\int \frac{4 x ^{2} + 26 x + 48}{( x + 5 ) ( x + 2 ) ^{2}} d x

What is (\partial)/(\partial y)(2e^xcos(yz)) ?
The answer to (\partial)/(\partial y)(2e^xcos(yz)) is -2e^xzsin(yz)