What is (\partial)/(\partial x)(2xsin(8x^2y)) ?

The answer to (\partial)/(\partial x)(2xsin(8x^2y)) is 2(sin(8x^2y)+16x^2ycos(8x^2y))

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(\partial)/(\partial x)(2xsin(8x^2y))

\frac{\partial}{\partial x} (2 x sin (8 x^{2} y))

2 (sin (8 x^{2} y) + 16 x^{2} y cos (8 x^{2} y))

Solution steps

\frac{\partial}{\partial x} (2 x sin (8 x^{2} y))

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

f = x, g = sin (8 x^{2} y)

= 2 (\frac{\partial}{\partial x} (x) sin (8 x^{2} y) + \frac{\partial}{\partial x} (sin (8 x^{2} y)) x)

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

\frac{\partial}{\partial x} (sin (8 x^{2} y)) = cos (8 x^{2} y) \cdot 16 y x

= 2 (1 \cdot sin (8 x^{2} y) + cos (8 x^{2} y) \cdot 16 y xx)

Simplify

2 (1 \cdot sin (8 x^{2} y) + cos (8 x^{2} y) \cdot 16 y xx) : 2 (sin (8 x^{2} y) + 16 x^{2} y cos (8 x^{2} y))

= 2 (sin (8 x^{2} y) + 16 x^{2} y cos (8 x^{2} y))

y^{^{'}} = 0.0012 y (1700 - y)

y^{^{'}} = \frac{( 2 x ^{2} + 3 y ^{2} )}{2 x y}

\frac{\partial}{\partial v} (tan (\frac{u}{v}))

n = 3 \sum \infty \frac{2}{n ( n + 2 )}

\int - 5 csc^{2} (x) d x

What is (\partial)/(\partial x)(2xsin(8x^2y)) ?
The answer to (\partial)/(\partial x)(2xsin(8x^2y)) is 2(sin(8x^2y)+16x^2ycos(8x^2y))