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

The answer to (\partial)/(\partial x)(9xsin(4x^2y)) is 9(sin(4x^2y)+8x^2ycos(4x^2y))

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

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

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

Solution steps

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

= 9 (\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

= 9 (1 \cdot sin (4 x^{2} y) + cos (4 x^{2} y) \cdot 8 y xx)

Simplify

9 (1 \cdot sin (4 x^{2} y) + cos (4 x^{2} y) \cdot 8 y xx) : 9 (sin (4 x^{2} y) + 8 x^{2} y cos (4 x^{2} y))

= 9 (sin (4 x^{2} y) + 8 x^{2} y cos (4 x^{2} y))

f (t) = t^{2}

\frac{d}{d x} (\frac{( 4 x ^{2} + 8 x + 2 )}{x})

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

\int x (x + 1)^{2} d x

\frac{d y}{d x} = x^{3} (x - x y)

What is (\partial)/(\partial x)(9xsin(4x^2y)) ?
The answer to (\partial)/(\partial x)(9xsin(4x^2y)) is 9(sin(4x^2y)+8x^2ycos(4x^2y))