What is (\partial)/(\partial x)(2xe^{3xy}) ?

The answer to (\partial)/(\partial x)(2xe^{3xy}) is 2(e^{3xy}+3e^{3xy}xy)

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(\partial)/(\partial x)(2xe^{3xy})

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

2 (e^{3 x y} + 3 e^{3 x y} x y)

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

f = x, g = e^{3 x y}

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

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

\frac{\partial}{\partial x} (e^{3 x y}) = e^{3 x y} \cdot 3 y

= 2 (1 \cdot e^{3 x y} + e^{3 x y} \cdot 3 y x)

Simplify

= 2 (e^{3 x y} + 3 e^{3 x y} x y)

(sin (x) cos (y)) d x + (cos (x) sin (y)) d y = 0

- x^{4} - 6 x^{3} + 8 x - 6

\frac{d}{d x} (x^{2} arctan (x^{3}))

x^{3} + 4 x^{2} + 7 x + 1

n = 2 \sum \infty \frac{6}{7 ^{n}}

What is (\partial)/(\partial x)(2xe^{3xy}) ?
The answer to (\partial)/(\partial x)(2xe^{3xy}) is 2(e^{3xy}+3e^{3xy}xy)