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

The answer to (\partial)/(\partial x)(3xsin(4x^2y)) is 3(sin(4x^2y)+8x^2ycos(4x^2y))

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

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

3 (sin (4 x^{2} y) + 8 x^{2} y cos (4 x^{2} y))

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

f = x, g = sin (4 x^{2} y)

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

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

\frac{\partial}{\partial x} (sin (4 x^{2} y)) = cos (4 x^{2} y) \cdot 8 y x

= 3 (1 \cdot sin (4 x^{2} y) + cos (4 x^{2} y) \cdot 8 y xx)

Simplify

3 (1 \cdot sin (4 x^{2} y) + cos (4 x^{2} y) \cdot 8 y xx) : 3 (sin (4 x^{2} y) + 8 x^{2} y cos (4 x^{2} y))

= 3 (sin (4 x^{2} y) + 8 x^{2} y cos (4 x^{2} y))

y = 3 (x + 1), y = 4 (x + 1), x = 3

\int_{0}^{\frac{3 π}{2}} 2 ∣ sin (x) ∣ d x

\int e^{\frac{y}{3}} d y

\frac{x ^{2} - 4 x + 2}{x + 3}

x \to 1 lim (x^{2} - 1)

What is (\partial)/(\partial x)(3xsin(4x^2y)) ?
The answer to (\partial)/(\partial x)(3xsin(4x^2y)) is 3(sin(4x^2y)+8x^2ycos(4x^2y))