What is (\partial)/(\partial x)(x*e^{x^3y}) ?

The answer to (\partial)/(\partial x)(x*e^{x^3y}) is e^{x^3y}+3e^{x^3y}x^3y

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(\partial)/(\partial x)(x*e^{x^3y})

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

e^{x^{3} y} + 3 e^{x^{3} y} x^{3} y

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

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

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

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

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

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

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

Simplify

1 \cdot e^{x^{3} y} + e^{x^{3} y} \cdot 3 y x^{2} x : e^{x^{3} y} + 3 e^{x^{3} y} x^{3} y

= e^{x^{3} y} + 3 e^{x^{3} y} x^{3} y

\int (\frac{1}{x} - 6 e^{x}) d x

\int \frac{3}{4 + x ^{2}} d x

y d y + (1 + y^{2}) sin (t) d t = 0

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

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

What is (\partial)/(\partial x)(x*e^{x^3y}) ?
The answer to (\partial)/(\partial x)(x*e^{x^3y}) is e^{x^3y}+3e^{x^3y}x^3y