What is the d/(dy)(ln(3x^2-y^3-3)) ?

The d/(dy)(ln(3x^2-y^3-3)) is -(3y^2)/(3x^2-y^3-3)

What is the first d/(dy)(ln(3x^2-y^3-3)) ?

The first d/(dy)(ln(3x^2-y^3-3)) is -(3y^2)/(3x^2-y^3-3)

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

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

Solution steps

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

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

\frac{d}{d 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{3 x ^{2}}{1 - x - x ^{5}}

x \to 3 lim (ln (\frac{1}{3 - x}) - 3)

\int \frac{x ^{2} + 6 x - 5}{2 x ^{3} + 3 x ^{2} - 2 x} d x

\frac{s + 1}{s ^{2} + 25}

\int 10 x e^{x^{2}} d x

What is the d/(dy)(ln(3x^2-y^3-3)) ?
The d/(dy)(ln(3x^2-y^3-3)) is -(3y^2)/(3x^2-y^3-3)
What is the first d/(dy)(ln(3x^2-y^3-3)) ?
The first d/(dy)(ln(3x^2-y^3-3)) is -(3y^2)/(3x^2-y^3-3)