What is the derivative of ln(3x^2+1ln(2x^3+1)) ?

The derivative of ln(3x^2+1ln(2x^3+1)) is (6xln(2x^3+1))/(3x^2+1)+(6x^2ln(3x^2+1))/(2x^3+1)

What is the first derivative of ln(3x^2+1ln(2x^3+1)) ?

The first derivative of ln(3x^2+1ln(2x^3+1)) is (6xln(2x^3+1))/(3x^2+1)+(6x^2ln(3x^2+1))/(2x^3+1)

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\frac{d}{d x} (ln (3 x^{2} + 1) ln (2 x^{3} + 1))

\frac{6 x ln ( 2 x ^{3} + 1 )}{3 x ^{2} + 1} + \frac{6 x ^{2} ln ( 3 x ^{2} + 1 )}{2 x ^{3} + 1}

Solution steps

\frac{d}{d x} (ln (3 x^{2} + 1) ln (2 x^{3} + 1))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = ln (3 x^{2} + 1), g = ln (2 x^{3} + 1)

= \frac{d}{d x} (ln (3 x^{2} + 1)) ln (2 x^{3} + 1) + \frac{d}{d x} (ln (2 x^{3} + 1)) ln (3 x^{2} + 1)

\frac{d}{d x} (ln (3 x^{2} + 1)) = \frac{6 x}{3 x ^{2} + 1}

\frac{d}{d x} (ln (2 x^{3} + 1)) = \frac{6 x ^{2}}{2 x ^{3} + 1}

= \frac{6 x}{3 x ^{2} + 1} ln (2 x^{3} + 1) + \frac{6 x ^{2}}{2 x ^{3} + 1} ln (3 x^{2} + 1)

Simplify

\frac{6 x}{3 x ^{2} + 1} ln (2 x^{3} + 1) + \frac{6 x ^{2}}{2 x ^{3} + 1} ln (3 x^{2} + 1) : \frac{6 x ln ( 2 x ^{3} + 1 )}{3 x ^{2} + 1} + \frac{6 x ^{2} ln ( 3 x ^{2} + 1 )}{2 x ^{3} + 1}

= \frac{6 x ln ( 2 x ^{3} + 1 )}{3 x ^{2} + 1} + \frac{6 x ^{2} ln ( 3 x ^{2} + 1 )}{2 x ^{3} + 1}

\int \frac{x ^{5} + x + 1}{x ^{3} - 8} d x

y = (x^{2} + 3) ((2 x + 5)^{6})

\frac{d}{d x} (5 + \frac{1}{x ^{2}})

\int_{0}^{3} x^{2} - 4 d x

\int 29 x^{28} d x

What is the derivative of ln(3x^2+1ln(2x^3+1)) ?
The derivative of ln(3x^2+1ln(2x^3+1)) is (6xln(2x^3+1))/(3x^2+1)+(6x^2ln(3x^2+1))/(2x^3+1)
What is the first derivative of ln(3x^2+1ln(2x^3+1)) ?
The first derivative of ln(3x^2+1ln(2x^3+1)) is (6xln(2x^3+1))/(3x^2+1)+(6x^2ln(3x^2+1))/(2x^3+1)