What is (\partial)/(\partial x)((x-1)y^2e^{xy}) ?

The answer to (\partial)/(\partial x)((x-1)y^2e^{xy}) is y^2(e^{xy}+e^{xy}y(x-1))

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(\partial)/(\partial x)((x-1)y^2e^{xy})

\frac{\partial}{\partial x} ((x - 1) y^{2} e^{x y})

y^{2} (e^{x y} + e^{x y} y (x - 1))

Solution steps

\frac{\partial}{\partial x} ((x - 1) y^{2} e^{x y})

Treat

y

as a constant

Take the constant out:

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

= y^{2} \frac{\partial}{\partial x} ((x - 1) e^{x y})

Apply the Product Rule:

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

f = x - 1, g = e^{x y}

= y^{2} (\frac{\partial}{\partial x} (x - 1) e^{x y} + \frac{\partial}{\partial x} (e^{x y}) (x - 1))

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

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

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

Simplify

= y^{2} (e^{x y} + e^{x y} y (x - 1))

\frac{d y}{d x} = \frac{y}{2 x}

n = 1 \sum \infty \frac{n !}{n ^{2} + 1}

n = 1 \sum \infty 0. 8^{n}

\int (sec^{2} (x)) d x

What is (\partial)/(\partial x)((x-1)y^2e^{xy}) ?
The answer to (\partial)/(\partial x)((x-1)y^2e^{xy}) is y^2(e^{xy}+e^{xy}y(x-1))