What is (\partial)/(\partial x)(sin(-3x)cos(4y)) ?

The answer to (\partial)/(\partial x)(sin(-3x)cos(4y)) is -cos(4y)cos(3x)*3

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(\partial)/(\partial x)(sin(-3x)cos(4y))

\frac{\partial}{\partial x} (sin (- 3 x) cos (4 y))

- cos (4 y) cos (3 x) \cdot 3

Solution steps

\frac{\partial}{\partial x} (sin (- 3 x) cos (4 y))

Simplify

sin (- 3 x) cos (4 y) : - sin (3 x) cos (4 y)

= \frac{\partial}{\partial x} (- sin (3 x) cos (4 y))

Treat

y

as a constant

Take the constant out:

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

= - cos (4 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 (4 y) cos (3 x) \cdot 3

\frac{d}{d x} (x^{3} - 6 x^{2} + 11 x - 6)

\int 1 - y^{2} d y

\int \frac{1}{f ( f - u )} df

\int_{1}^{5} (4 x - \frac{4}{x}) d x

What is (\partial)/(\partial x)(sin(-3x)cos(4y)) ?
The answer to (\partial)/(\partial x)(sin(-3x)cos(4y)) is -cos(4y)cos(3x)*3