What is (\partial)/(\partial x)(4xsin(3x^2y)) ?

The answer to (\partial)/(\partial x)(4xsin(3x^2y)) is 4(sin(3x^2y)+6x^2ycos(3x^2y))

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(\partial)/(\partial x)(4xsin(3x^2y))

\frac{\partial}{\partial x} (4 x sin (3 x^{2} y))

4 (sin (3 x^{2} y) + 6 x^{2} y cos (3 x^{2} y))

Solution steps

\frac{\partial}{\partial x} (4 x sin (3 x^{2} y))

Treat

y

as a constant

Take the constant out:

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

= 4 \frac{\partial}{\partial x} (x sin (3 x^{2} y))

Apply the Product Rule:

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

f = x, g = sin (3 x^{2} y)

= 4 (\frac{\partial}{\partial x} (x) sin (3 x^{2} y) + \frac{\partial}{\partial x} (sin (3 x^{2} y)) x)

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

\frac{\partial}{\partial x} (sin (3 x^{2} y)) = cos (3 x^{2} y) \cdot 6 y x

= 4 (1 \cdot sin (3 x^{2} y) + cos (3 x^{2} y) \cdot 6 y xx)

Simplify

4 (1 \cdot sin (3 x^{2} y) + cos (3 x^{2} y) \cdot 6 y xx) : 4 (sin (3 x^{2} y) + 6 x^{2} y cos (3 x^{2} y))

= 4 (sin (3 x^{2} y) + 6 x^{2} y cos (3 x^{2} y))

\int_{- 2}^{18} \frac{1}{∣ 2 x ∣} d x

\frac{d}{d x} ((4 x^{3} + 6) (x + 8))

x \to \infty lim (7 x^{2} + 8)

x \to 0 + lim (sin (x) \cdot ln (2 x))

\frac{d}{d t} (e^{- t} cos (2 t))

What is (\partial)/(\partial x)(4xsin(3x^2y)) ?
The answer to (\partial)/(\partial x)(4xsin(3x^2y)) is 4(sin(3x^2y)+6x^2ycos(3x^2y))