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

The answer to (\partial)/(\partial y)(msin(2(mx+ny))) is mcos(2(mx+ny))*2n

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

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

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

Solution steps

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

Treat

n, m, x

as constants

Take the constant out:

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

= m \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

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

y = \frac{1}{3 t}, (2, \frac{1}{6})

\int x \cdot cos^{2} (x) d x

g (x) = \frac{x}{x - 1}

f^{^{''}} (x) = \frac{6}{x}

x \to 0 lim (1 - 2 sin^{2} (\frac{x}{2}))

What is (\partial)/(\partial y)(msin(2(mx+ny))) ?
The answer to (\partial)/(\partial y)(msin(2(mx+ny))) is mcos(2(mx+ny))*2n