What is (\partial)/(\partial y)(sin^2(mx+ny)) ?

The answer to (\partial)/(\partial y)(sin^2(mx+ny)) is nsin(2(mx+ny))

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(\partial)/(\partial y)(sin^2(mx+ny))

\frac{\partial}{\partial y} (sin^{2} (m x + n y))

n sin (2 (m x + n y))

Solution steps

\frac{\partial}{\partial y} (sin^{2} (m x + n y))

Treat

n, m, x

as constants

Apply the chain rule:

2 sin (m x + n y) \frac{\partial}{\partial y} (sin (m x + n y))

= 2 sin (m x + n y) \frac{\partial}{\partial y} (sin (m x + n y))

\frac{\partial}{\partial y} (sin (m x + n y)) = cos (m x + n y) n

= 2 sin (m x + n y) cos (m x + n y) n

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

= n sin (2 (m x + n y))

\frac{d}{d x} (ln (x^{5} - 3 x^{3}))

n = 0 \sum \infty \frac{1}{6 ^{n}}

x y^{^{'}} + 2 y = 12 x^{2}

\frac{d}{d t} (sin (t^{3}))

\int \frac{x ^{3} + 3 x ^{2} - 5 x - 1}{x ^{2} - x} d x

What is (\partial)/(\partial y)(sin^2(mx+ny)) ?
The answer to (\partial)/(\partial y)(sin^2(mx+ny)) is nsin(2(mx+ny))