What is (\partial)/(\partial x)(2xsin(9x^2y)) ?

The answer to (\partial)/(\partial x)(2xsin(9x^2y)) is 2(sin(9x^2y)+18x^2ycos(9x^2y))

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(\partial)/(\partial x)(2xsin(9x^2y))

\frac{\partial}{\partial x} (2 x sin (9 x^{2} y))

2 (sin (9 x^{2} y) + 18 x^{2} y cos (9 x^{2} y))

Solution steps

\frac{\partial}{\partial x} (2 x sin (9 x^{2} y))

Treat

y

as a constant

Take the constant out:

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

= 2 \frac{\partial}{\partial x} (x sin (9 x^{2} y))

Apply the Product Rule:

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

f = x, g = sin (9 x^{2} y)

= 2 (\frac{\partial}{\partial x} (x) sin (9 x^{2} y) + \frac{\partial}{\partial x} (sin (9 x^{2} y)) x)

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

\frac{\partial}{\partial x} (sin (9 x^{2} y)) = cos (9 x^{2} y) \cdot 18 y x

= 2 (1 \cdot sin (9 x^{2} y) + cos (9 x^{2} y) \cdot 18 y xx)

Simplify

2 (1 \cdot sin (9 x^{2} y) + cos (9 x^{2} y) \cdot 18 y xx) : 2 (sin (9 x^{2} y) + 18 x^{2} y cos (9 x^{2} y))

= 2 (sin (9 x^{2} y) + 18 x^{2} y cos (9 x^{2} y))

\int x sin (5 x) d x

\frac{d}{d x} (arcsin (sin (7 x)))

y = arcsec (\frac{1}{7 t ^{3}})

y = ∣ x - 3 ∣, y = \frac{x}{2}

f (x) = tan (sec (x))

What is (\partial)/(\partial x)(2xsin(9x^2y)) ?
The answer to (\partial)/(\partial x)(2xsin(9x^2y)) is 2(sin(9x^2y)+18x^2ycos(9x^2y))