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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial u ( v )} (u (v)^{2} e^{- v})

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

e^{7 t}

\frac{d}{d x} (ln (x^{2} - 7 x + 1))

\int \frac{1}{9 x ^{2} + 12 x + 8} d x

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