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)

\int (\frac{1}{2 x}) d x

\frac{d}{d x} (F (x^{4}))

f (x) = 2 x^{2} - x + 1

\int - 8 x^{2} d x

\frac{e ^{- π (s)}}{s ^{3} + s}

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