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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

f (h) = e^{(3 h^{2} + 6)}

\frac{\partial}{\partial x} (\frac{1}{ln ( \frac{x}{2} )})

y = e^{5 x}, at x = \frac{1}{5} ln (3)

(- 2 sin (x))^{^{'}}

x^{3} + \frac{1}{x ^{3}}

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