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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{x e ^{2 x} + 4 e ^{x}}{x e ^{x}} d x

y = 4 x + 26, (- 1, 5)

v (t) = t^{2} e^{- 3 t}

\frac{d y}{d t} = \frac{t}{y 1 + t ^{2}}

\frac{5}{s ( s + 1 ) ( s + 2 )}

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