What is (\partial)/(\partial x)(y/(2sqrt(x))) ?

The answer to (\partial)/(\partial x)(y/(2sqrt(x))) is -y/(4x^{3/2)}

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(\partial)/(\partial x)(y/(2sqrt(x)))

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

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

Simplify

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

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

x \cdot sin (\frac{4}{x})

x^{^{''}} - 8 x^{^{'}} + 16 x = 4 t e^{4 t}

\int - cos (4 x) d x

\frac{d}{d x} (7^{x + a r c c o s (x)})

y^{^{''}} + 4 y^{^{'}} + 20 y = 0

What is (\partial)/(\partial x)(y/(2sqrt(x))) ?
The answer to (\partial)/(\partial x)(y/(2sqrt(x))) is -y/(4x^{3/2)}