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)

\frac{d}{d x} (1 + x^{2} + x^{2} 1 + x^{2})

\int (5 e^{2 x} + \frac{1}{x}) d x

\frac{d}{d x} (sin (k t))

f (x) = (5 + 2 x) e^{- 3 x}

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

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