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

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

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

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

- 3 sin (3 x) sin (3 y)

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

- sin (3 y) \frac{\partial}{\partial y} (3 y)

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

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

= sin (3 x) (- sin (3 y) \cdot 3)

Simplify

= - 3 sin (3 x) sin (3 y)

\frac{( e ^{\frac{b}{x}} )}{( e ^{\frac{b}{x}} - 1 )}

\frac{d}{d x} ((x^{2} sin (x))^{5})

\int \frac{e ^{x}}{x + 1} d x

\frac{d}{d t} (\frac{sec ^{2} ( t )}{sec ( t ) tan ( t )})

\int_{- 2}^{1} \frac{1}{x ^{4}} d x

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