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

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

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

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

- \frac{3 y ^{2}}{3 x ^{2} - y ^{3} - 3}

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

= - \frac{3 y ^{2}}{3 x ^{2} - y ^{3} - 3}

\frac{d}{d x} (16 I n (x^{2}) - x^{2})

\int sin (2 x) cos^{2} (x) d x

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

\int x (1 - x)^{99} d x

\int \frac{x}{x ^{4} + 4 x ^{2} + 5} d x

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