What is the derivative of log_{3}(x^3+1) ?

The derivative of log_{3}(x^3+1) is (3x^2)/(ln(3)(x^3+1))

What is the first derivative of log_{3}(x^3+1) ?

The first derivative of log_{3}(x^3+1) is (3x^2)/(ln(3)(x^3+1))

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derivative of

lo g_{3} (x^{3} + 1)

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

Solution steps

\frac{d}{d x} (lo g_{3} (x^{3} + 1))

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{x}{3})

a re a y = 2 x, [4, 9]

\frac{d}{d x} ((3 x^{2} + 2 x - 1)^{4})

\frac{d y}{d t} = (y^{2} + 1) t

\frac{d}{d x} (10 + x)

What is the derivative of log_{3}(x^3+1) ?
The derivative of log_{3}(x^3+1) is (3x^2)/(ln(3)(x^3+1))
What is the first derivative of log_{3}(x^3+1) ?
The first derivative of log_{3}(x^3+1) is (3x^2)/(ln(3)(x^3+1))