What is (\partial)/(\partial x)(2xsin(xy)) ?

The answer to (\partial)/(\partial x)(2xsin(xy)) is 2(sin(xy)+xycos(xy))

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(\partial)/(\partial x)(2xsin(xy))

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

2 (sin (x y) + x y cos (x y))

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = sin (x y)

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

\frac{\partial}{\partial x} (x) = 1

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

= 2 (1 \cdot sin (x y) + cos (x y) y x)

Simplify

= 2 (sin (x y) + x y cos (x y))

x v^{'} + 2 v = 0

x y^{^{'}} = 2 y + x^{3} cos (x)

\int cos^{5} (x) sin^{4} (x) d x

\frac{d y}{d x} 0.4 x y = 9 x

\int \frac{2 x - 11}{x ^{2} ( x + 2 )} d x

What is (\partial)/(\partial x)(2xsin(xy)) ?
The answer to (\partial)/(\partial x)(2xsin(xy)) is 2(sin(xy)+xycos(xy))