What is (\partial)/(\partial x)((x^4+y^2)/(xy)) ?

The answer to (\partial)/(\partial x)((x^4+y^2)/(xy)) is (3x^4-y^2)/(x^2y)

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(\partial)/(\partial x)((x^4+y^2)/(xy))

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

\frac{3 x ^{4} - y ^{2}}{x ^{2} y}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= \frac{1}{y} \frac{\partial}{\partial x} (\frac{x ^{4} + y ^{2}}{x})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{y} \cdot \frac{\frac{\partial}{\partial x} ( x ^{4} + y ^{2} ) x - \frac{\partial}{\partial x} ( x ) ( x ^{4} + y ^{2} )}{x ^{2}}

\frac{\partial}{\partial x} (x^{4} + y^{2}) = 4 x^{3}

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

= \frac{1}{y} \cdot \frac{4 x ^{3} x - 1 \cdot ( x ^{4} + y ^{2} )}{x ^{2}}

Simplify

\frac{1}{y} \cdot \frac{4 x ^{3} x - 1 \cdot ( x ^{4} + y ^{2} )}{x ^{2}} : \frac{3 x ^{4} - y ^{2}}{x ^{2} y}

= \frac{3 x ^{4} - y ^{2}}{x ^{2} y}

\int 9 x + 5 d x

x \to - 6 + lim (\frac{- 2 x}{x + 6})

(0.2) (2)

\frac{d}{d x} (x^{10} e^{x + 1})

\frac{d y}{d t} = y (1 - y), y (0) = 16

What is (\partial)/(\partial x)((x^4+y^2)/(xy)) ?
The answer to (\partial)/(\partial x)((x^4+y^2)/(xy)) is (3x^4-y^2)/(x^2y)