What is (\partial)/(\partial y)(1/2 sin(2x-y)) ?

The answer to (\partial)/(\partial y)(1/2 sin(2x-y)) is -1/2 cos(2x-y)

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(\partial)/(\partial y)(1/2 sin(2x-y))

\frac{\partial}{\partial y} (\frac{1}{2} sin (2 x - y))

- \frac{1}{2} cos (2 x - y)

Solution steps

\frac{\partial}{\partial y} (\frac{1}{2} sin (2 x - y))

Treat

x

as a constant

Take the constant out:

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

= \frac{1}{2} \frac{\partial}{\partial y} (sin (2 x - y))

Apply the chain rule:

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

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

\frac{\partial}{\partial y} (2 x - y) = - 1

= \frac{1}{2} cos (2 x - y) (- 1)

Simplify

= - \frac{1}{2} cos (2 x - y)

\int_{5}^{\infty} x e^{- 2 x} d x

f (x) = 7 x^{2} + 4 x

\int 3 e^{x} + 8 x^{3} + 4 x d x

\frac{d}{d x} (cos (\frac{1}{4} x^{3}))

\frac{d}{d x} (\frac{e ^{4 x}}{x ^{8}})

What is (\partial)/(\partial y)(1/2 sin(2x-y)) ?
The answer to (\partial)/(\partial y)(1/2 sin(2x-y)) is -1/2 cos(2x-y)