What is (\partial)/(\partial y)((x^2)/(1+y^3)) ?

The answer to (\partial)/(\partial y)((x^2)/(1+y^3)) is -(3x^2y^2)/((1+y^3)^2)

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(\partial)/(\partial y)((x^2)/(1+y^3))

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (a x e^{b x^{2}})

\int - \frac{arccot ( x )}{1 + x ^{2}} d x

\frac{\partial}{\partial x} (3 x^{2} + 6 x y)

\frac{d}{d x} (\frac{12 ( 3 x ^{3} - 16 )}{4 x - x ^{4}})

\int p (p + 7)^{8} d p

What is (\partial)/(\partial y)((x^2)/(1+y^3)) ?
The answer to (\partial)/(\partial y)((x^2)/(1+y^3)) is -(3x^2y^2)/((1+y^3)^2)