What is (\partial)/(\partial y)(xyz+sin(2xy)) ?

The answer to (\partial)/(\partial y)(xyz+sin(2xy)) is xz+cos(2xy)*2x

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(\partial)/(\partial y)(xyz+sin(2xy))

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

x z + cos (2 x y) \cdot 2 x

Solution steps

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

Treat

x, z

as constants

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial y} (x yz) = x z

\frac{\partial}{\partial y} (sin (2 x y)) = cos (2 x y) \cdot 2 x

= x z + cos (2 x y) \cdot 2 x

5 cos (π x), 12 x^{3} - 3, 0.5, - 0.5

\frac{1}{1 + x}, 1

f (x) = x^{3} - 6 x, g (x) = - 2 x

36 t^{2}

\int x^{2} 3 x + 6 d x

What is (\partial)/(\partial y)(xyz+sin(2xy)) ?
The answer to (\partial)/(\partial y)(xyz+sin(2xy)) is xz+cos(2xy)*2x