What is (\partial)/(\partial x)(1/(xy)(x^2y^2+y+x)) ?

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

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(\partial)/(\partial x)(1/(xy)(x^2y^2+y+x))

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

\frac{x ^{2} y - 1}{x ^{2}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

f = \frac{1}{x}, g = x^{2} y^{2} + y + x

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

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

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

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

Simplify

\frac{1}{y} ((- \frac{1}{x ^{2}}) (x^{2} y^{2} + y + x) + (2 y^{2} x + 1) \frac{1}{x}) : \frac{x ^{2} y - 1}{x ^{2}}

= \frac{x ^{2} y - 1}{x ^{2}}

\frac{\partial}{\partial x} (5 x y^{4} + 2 y cos (x))

y = x^{3} e^{4 x}

\int \frac{e ^{- 2 x} + e ^{2 x}}{2} d x

f (x) = \frac{18}{( 2 x + 4 ) ^{2}}, at x = 9

\int cos (2 x) e^{2 x} d x

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