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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (tan (2 x^{3}))

\int \frac{x}{x ^{4} - 16 x ^{2} + 39} d x

y = x + x^{2} + x^{3}, x = 1

\int_{0}^{e} a^{x} d x

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

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