What is (\partial)/(\partial x)(4xsin(9x^2y)) ?

The answer to (\partial)/(\partial x)(4xsin(9x^2y)) is 4(sin(9x^2y)+18x^2ycos(9x^2y))

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(\partial)/(\partial x)(4xsin(9x^2y))

\frac{\partial}{\partial x} (4 x sin (9 x^{2} y))

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

Solution steps

\frac{\partial}{\partial x} (4 x sin (9 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 (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)

= 4 (\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

= 4 (1 \cdot sin (9 x^{2} y) + cos (9 x^{2} y) \cdot 18 y xx)

Simplify

4 (1 \cdot sin (9 x^{2} y) + cos (9 x^{2} y) \cdot 18 y xx) : 4 (sin (9 x^{2} y) + 18 x^{2} y cos (9 x^{2} y))

= 4 (sin (9 x^{2} y) + 18 x^{2} y cos (9 x^{2} y))

\frac{d}{d x} (arccos (\frac{x}{3}))

\frac{d}{d x} (3 x^{2} - 1)

\int_{0}^{2} (6 - 4 ln (3 - x))^{2} d x

\int 9 x d x

y y^{^{'}} = t

What is (\partial)/(\partial x)(4xsin(9x^2y)) ?
The answer to (\partial)/(\partial x)(4xsin(9x^2y)) is 4(sin(9x^2y)+18x^2ycos(9x^2y))