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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

y = (x - 2)^{\frac{2}{3}}, (3, 1)

\int e^{- \frac{1}{3} x} d x

\frac{d}{d x} (cot (e^{x}))

\int \frac{x + 4}{x ^{2} + 6 x + 18} d x

n = 3 \sum \infty (\frac{4}{10})^{n}

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