What is (\partial)/(\partial y)((e^x-4x)cos(y)) ?

The answer to (\partial)/(\partial y)((e^x-4x)cos(y)) is -sin(y)(e^x-4x)

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(\partial)/(\partial y)((e^x-4x)cos(y))

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

- sin (y) (e^{x} - 4 x)

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the common derivative:

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

= (e^{x} - 4 x) (- sin (y))

Simplify

= - sin (y) (e^{x} - 4 x)

\int \frac{1}{x} + \frac{y ^{2}}{x ^{2}} d x

\frac{d}{d x} (ln ((x - 1)^{3} (x^{2} + 1)^{4}))

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

\frac{d y}{d t} = - \frac{3 y}{t} - 2 - t^{- 4}

\frac{\partial}{\partial z} (cos (x y) + sin (yz))

What is (\partial)/(\partial y)((e^x-4x)cos(y)) ?
The answer to (\partial)/(\partial y)((e^x-4x)cos(y)) is -sin(y)(e^x-4x)