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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int_{2}^{e} \frac{1}{x ln ( x )} d x

f (x) = x (5 x^{5} - 1)

\int x sin (9 x^{2}) d x

\frac{d}{d x} (\frac{x ^{6}}{6} + \frac{1}{16 x ^{4}})

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