What is (\partial)/(\partial y)(sin^2(x-3y)) ?

The answer to (\partial)/(\partial y)(sin^2(x-3y)) is -6sin(x-3y)cos(x-3y)

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(\partial)/(\partial y)(sin^2(x-3y))

\frac{\partial}{\partial y} (sin^{2} (x - 3 y))

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

Solution steps

\frac{\partial}{\partial y} (sin^{2} (x - 3 y))

Treat

x

as a constant

Apply the chain rule:

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

= 2 sin (x - 3 y) \frac{\partial}{\partial y} (sin (x - 3 y))

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

= 2 sin (x - 3 y) (- 3 cos (x - 3 y))

Simplify

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

\frac{d}{d x} (s cos (x))

n = 2 \sum \infty \frac{1}{n ^{4}}

\frac{\partial}{\partial y} (arctan (9 x - 9 y))

\int sin (x) cos (3 x) d x

What is (\partial)/(\partial y)(sin^2(x-3y)) ?
The answer to (\partial)/(\partial y)(sin^2(x-3y)) is -6sin(x-3y)cos(x-3y)