What is (\partial)/(\partial y)(2cos(y)) ?

The answer to (\partial)/(\partial y)(2cos(y)) is -2sin(y)

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(\partial)/(\partial y)(2cos(y))

\frac{\partial}{\partial y} (2 cos (y))

- 2 sin (y)

Solution steps

\frac{\partial}{\partial y} (2 cos (y))

Take the constant out:

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

= 2 \frac{\partial}{\partial y} (cos (y))

Apply the common derivative:

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

= 2 (- sin (y))

Simplify

= - 2 sin (y)

\int 2 e^{θ} d θ

(\frac{1}{x ^{2}} - \frac{3}{x ^{4}}) (x + 5 x^{3})

f (x) = \frac{2}{x - 4}

\int \frac{4 x ^{5} - 2 x + 3}{x ^{3}} d x

e^{- 2 s} (\frac{1}{s - 9})

What is (\partial)/(\partial y)(2cos(y)) ?
The answer to (\partial)/(\partial y)(2cos(y)) is -2sin(y)