What is (\partial)/(\partial x)(sin(-3x)cos(6y)) ?

The answer to (\partial)/(\partial x)(sin(-3x)cos(6y)) is -cos(6y)cos(3x)*3

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(\partial)/(\partial x)(sin(-3x)cos(6y))

\frac{\partial}{\partial x} (sin (- 3 x) cos (6 y))

- cos (6 y) cos (3 x) \cdot 3

Solution steps

\frac{\partial}{\partial x} (sin (- 3 x) cos (6 y))

Simplify

sin (- 3 x) cos (6 y) : - sin (3 x) cos (6 y)

= \frac{\partial}{\partial x} (- sin (3 x) cos (6 y))

Treat

y

as a constant

Take the constant out:

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

= - cos (6 y) \frac{\partial}{\partial x} (sin (3 x))

Apply the chain rule:

cos (3 x) \frac{\partial}{\partial x} (3 x)

= cos (3 x) \frac{\partial}{\partial x} (3 x)

\frac{\partial}{\partial x} (3 x) = 3

= - cos (6 y) cos (3 x) \cdot 3

\frac{d}{d x} ((cos (x))^{5})

x \to 0 lim (arcsin (x))

\frac{d}{d x} (lo g_{2} (x^{2} - 9 x))

\frac{d}{d x} ((x^{2} - 1) (x^{2} - 4) (x^{2} - 9))

\frac{\partial}{\partial x} (50 - (4 x^{2} + y^{2}))

What is (\partial)/(\partial x)(sin(-3x)cos(6y)) ?
The answer to (\partial)/(\partial x)(sin(-3x)cos(6y)) is -cos(6y)cos(3x)*3