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

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

(\partial)/(\partial y)(nsin(2(mx+ny)))

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

2 n^{2} cos (2 (m x + n y))

Solution steps

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

Treat

n, m, x

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= n cos (2 (m x + n y)) \cdot 2 n

Simplify

n cos (2 (m x + n y)) \cdot 2 n : 2 n^{2} cos (2 (m x + n y))

= 2 n^{2} cos (2 (m x + n y))

in v erse l a pl a ce \frac{30}{( 10 s + 1 ) ( s ^{2} + 4 )}

d er i v a t i v e x^{x + 2}

\frac{d}{d x} (\frac{x ^{2} - 64}{x - 8})

d er i v a t i v e f (x) = 3 - x

x \to - 1 lim (π)

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