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)

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(\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}}

\frac{5}{( 2 x - 3 ) ^{4}}

\frac{1 - x}{1 + x}

\int 2 x \cdot sin (x) d x

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

x, 6 - x, [4, 9]

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)