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

The answer to (\partial)/(\partial x)(y/(2sqrt(1+xy))) is -(y^2)/(4(1+yx)^{3/2)}

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

- \frac{1}{2 ( 1 + x y ) ^{\frac{3}{2}}} \frac{\partial}{\partial x} (1 + x y)

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

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

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

Simplify

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

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

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What is (\partial)/(\partial x)(y/(2sqrt(1+xy))) ?
The answer to (\partial)/(\partial x)(y/(2sqrt(1+xy))) is -(y^2)/(4(1+yx)^{3/2)}