What is (\partial)/(\partial y)((6x)/(3x^2-y^3-3)) ?

The answer to (\partial)/(\partial y)((6x)/(3x^2-y^3-3)) is (18xy^2)/((3x^2-y^3-3)^2)

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(\partial)/(\partial y)((6x)/(3x^2-y^3-3))

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

\frac{18 x y ^{2}}{( 3 x ^{2} - y ^{3} - 3 ) ^{2}}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

= 6 x \frac{\partial}{\partial y} (\frac{1}{3 x ^{2} - y ^{3} - 3})

Apply exponent rule:

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

= 6 x \frac{\partial}{\partial y} ((3 x^{2} - y^{3} - 3)^{- 1})

Apply the chain rule:

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

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

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

= 6 x (- \frac{1}{( 3 x ^{2} - y ^{3} - 3 ) ^{2}} (- 3 y^{2}))

Simplify

6 x (- \frac{1}{( 3 x ^{2} - y ^{3} - 3 ) ^{2}} (- 3 y^{2})) : \frac{18 x y ^{2}}{( 3 x ^{2} - y ^{3} - 3 ) ^{2}}

= \frac{18 x y ^{2}}{( 3 x ^{2} - y ^{3} - 3 ) ^{2}}

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

x^{7} - 3 x^{4} + 7 x^{3} + 43, at x = - 2

\int \frac{1}{x 1 + x} d x

x \to 0 lim (3 - 4 x)

\int x^{2} + 81 d x

What is (\partial)/(\partial y)((6x)/(3x^2-y^3-3)) ?
The answer to (\partial)/(\partial y)((6x)/(3x^2-y^3-3)) is (18xy^2)/((3x^2-y^3-3)^2)