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

The answer to (\partial)/(\partial x)(msin(2mx+2ny)) is 2m^2cos(2mx+2ny)

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

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

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

Solution steps

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

Treat

m, n, y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int_{0}^{1} 2 x - x^{2} d x

\frac{\partial}{\partial x} (5 sin (x y))

\frac{d}{d x} (\frac{x ^{2}}{3 + x})

y^{^{'}} + 6 x y^{2} = 0

\frac{d}{d x} (\frac{1000}{x})

What is (\partial)/(\partial x)(msin(2mx+2ny)) ?
The answer to (\partial)/(\partial x)(msin(2mx+2ny)) is 2m^2cos(2mx+2ny)