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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\int_{0}^{0.5} 12 x - \frac{1}{12} x d x

\int cos^{5} (9 - 11 x) d x

\frac{d}{d x} (\frac{3 x - 4}{2 x - 8})

t^{2} y^{^{''}} - 3 t y^{^{'}} + 4 y = t^{2} \cdot ln (t)

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