What is (\partial)/(\partial x)(9xsin(6x^2y)) ?

The answer to (\partial)/(\partial x)(9xsin(6x^2y)) is 9(sin(6x^2y)+12x^2ycos(6x^2y))

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(\partial)/(\partial x)(9xsin(6x^2y))

\frac{\partial}{\partial x} (9 x sin (6 x^{2} y))

9 (sin (6 x^{2} y) + 12 x^{2} y cos (6 x^{2} y))

Solution steps

\frac{\partial}{\partial x} (9 x sin (6 x^{2} y))

Treat

y

as a constant

Take the constant out:

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

= 9 \frac{\partial}{\partial x} (x sin (6 x^{2} y))

Apply the Product Rule:

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

f = x, g = sin (6 x^{2} y)

= 9 (\frac{\partial}{\partial x} (x) sin (6 x^{2} y) + \frac{\partial}{\partial x} (sin (6 x^{2} y)) x)

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

\frac{\partial}{\partial x} (sin (6 x^{2} y)) = cos (6 x^{2} y) \cdot 12 y x

= 9 (1 \cdot sin (6 x^{2} y) + cos (6 x^{2} y) \cdot 12 y xx)

Simplify

9 (1 \cdot sin (6 x^{2} y) + cos (6 x^{2} y) \cdot 12 y xx) : 9 (sin (6 x^{2} y) + 12 x^{2} y cos (6 x^{2} y))

= 9 (sin (6 x^{2} y) + 12 x^{2} y cos (6 x^{2} y))

y = \frac{5 x}{1 + x ^{2}}, (5, \frac{25}{26})

f (x) = (\frac{x + 5}{x ^{2} + 2})^{2}

1 0^{x^{2} - 1}

\int w csc (4 w^{2} - 1) d w

x^{2} \frac{d ^{2} y}{d x ^{2}} = 0

What is (\partial)/(\partial x)(9xsin(6x^2y)) ?
The answer to (\partial)/(\partial x)(9xsin(6x^2y)) is 9(sin(6x^2y)+12x^2ycos(6x^2y))