What is (\partial)/(\partial y)(sin(-2x)cos(6y)) ?

The answer to (\partial)/(\partial y)(sin(-2x)cos(6y)) is 6sin(2x)sin(6y)

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(\partial)/(\partial y)(sin(-2x)cos(6y))

\frac{\partial}{\partial y} (sin (- 2 x) cos (6 y))

6 sin (2 x) sin (6 y)

Solution steps

\frac{\partial}{\partial y} (sin (- 2 x) cos (6 y))

Simplify

sin (- 2 x) cos (6 y) : - sin (2 x) cos (6 y)

= \frac{\partial}{\partial y} (- sin (2 x) cos (6 y))

Treat

x

as a constant

Take the constant out:

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

= - sin (2 x) \frac{\partial}{\partial y} (cos (6 y))

Apply the chain rule:

- sin (6 y) \frac{\partial}{\partial y} (6 y)

= - sin (6 y) \frac{\partial}{\partial y} (6 y)

\frac{\partial}{\partial y} (6 y) = 6

= - sin (2 x) (- sin (6 y) \cdot 6)

Simplify

= 6 sin (2 x) sin (6 y)

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\frac{d}{d x} (tan (e^{π}))

\frac{d}{d x} (\frac{1}{2 x + 6})

\int \frac{1}{4 x ^{2}} d x

\int \frac{- 8 x ^{2}}{4 x - x ^{2}} d x

What is (\partial)/(\partial y)(sin(-2x)cos(6y)) ?
The answer to (\partial)/(\partial y)(sin(-2x)cos(6y)) is 6sin(2x)sin(6y)