What is (\partial)/(\partial y)(-e^{-x}zcos(y)) ?

The answer to (\partial)/(\partial y)(-e^{-x}zcos(y)) is e^{-x}zsin(y)

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(\partial)/(\partial y)(-e^{-x}zcos(y))

\frac{\partial}{\partial y} (- e^{- x} z cos (y))

e^{- x} z sin (y)

Solution steps

\frac{\partial}{\partial y} (- e^{- x} z cos (y))

Treat

x, z

as constants

Take the constant out:

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

= - e^{- x} z \frac{\partial}{\partial y} (cos (y))

Apply the common derivative:

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

= - e^{- x} z (- sin (y))

Simplify

= e^{- x} z sin (y)

(x^{2} \cdot 2^{x})^{7}

\frac{1}{5 u ^{5}}

4 (x - \frac{1}{x})^{4}, at x = 2

\int sin (θ) cos^{3} (θ) d θ

x \to 1 lim (x^{3} - 5 x)

What is (\partial)/(\partial y)(-e^{-x}zcos(y)) ?
The answer to (\partial)/(\partial y)(-e^{-x}zcos(y)) is e^{-x}zsin(y)