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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{e ^{x}}{4 + 9 e ^{2 x}} d x

- x^{2} + 26 x + 920

y^{^{'}} = \frac{( 1 + x )}{x y ^{6}}

\frac{d}{d x} (\frac{4 x - 1 + x}{x ^{2} - 1})

f (x) = x^{4} - 5 x^{2} + 2 x - 4, (1, 0)

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