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)

\frac{\partial}{\partial x} (x^{4} - y^{2})

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

f (x) = cos^{3} (x)

ln (\frac{1}{x})

\int e^{3 x + 2} d x

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