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

The answer to (\partial)/(\partial y)(ln(2x^3+3y^2)) is (6y)/(2x^3+3y^2)

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

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

\frac{6 y}{2 x ^{3} + 3 y ^{2}}

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

= \frac{1}{2 x ^{3} + 3 y ^{2}} \cdot 6 y

Simplify

\frac{1}{2 x ^{3} + 3 y ^{2}} \cdot 6 y : \frac{6 y}{2 x ^{3} + 3 y ^{2}}

= \frac{6 y}{2 x ^{3} + 3 y ^{2}}

What is (\partial)/(\partial y)(ln(2x^3+3y^2)) ?
The answer to (\partial)/(\partial y)(ln(2x^3+3y^2)) is (6y)/(2x^3+3y^2)