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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

ln (3 x + 1), 3

\frac{d}{d x} (4 x^{2} + 5)

\frac{1}{( t ^{2} + 3 t + 1 ) ^{\frac{5}{2}}}

y = x^{4} + 8 x^{2} - x

\int 2 x^{3} - \frac{2}{3} x^{2} + 7 x d x

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