What is the derivative of log_{3}(3x^3+4x+4) ?

The derivative of log_{3}(3x^3+4x+4) is (9x^2+4)/(ln(3)(3x^3+4x+4))

What is the first derivative of log_{3}(3x^3+4x+4) ?

The first derivative of log_{3}(3x^3+4x+4) is (9x^2+4)/(ln(3)(3x^3+4x+4))

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derivative of

lo g_{3} (3 x^{3} + 4 x + 4)

\frac{9 x ^{2} + 4}{ln ( 3 ) ( 3 x ^{3} + 4 x + 4 )}

Solution steps

\frac{d}{d x} (lo g_{3} (3 x^{3} + 4 x + 4))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 3 x ^{3} + 4 x + 4 )}{ln ( 3 )})

Take the constant out:

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

= \frac{1}{ln ( 3 )} \frac{d}{d x} (ln (3 x^{3} + 4 x + 4))

Apply the chain rule:

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

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

\frac{d}{d x} (3 x^{3} + 4 x + 4) = 9 x^{2} + 4

= \frac{1}{ln ( 3 )} \cdot \frac{1}{3 x ^{3} + 4 x + 4} (9 x^{2} + 4)

Simplify

\frac{1}{ln ( 3 )} \cdot \frac{1}{3 x ^{3} + 4 x + 4} (9 x^{2} + 4) : \frac{9 x ^{2} + 4}{ln ( 3 ) ( 3 x ^{3} + 4 x + 4 )}

= \frac{9 x ^{2} + 4}{ln ( 3 ) ( 3 x ^{3} + 4 x + 4 )}

\int 7 sin^{6} (x) cos (x) d x

\int \frac{1}{e ^{2 y} - 2 e ^{y}} d y

f (x) = 6^{x} lo g_{5} (x)

\int_{0}^{2 π} cos^{2} (x) d x

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

What is the derivative of log_{3}(3x^3+4x+4) ?
The derivative of log_{3}(3x^3+4x+4) is (9x^2+4)/(ln(3)(3x^3+4x+4))
What is the first derivative of log_{3}(3x^3+4x+4) ?
The first derivative of log_{3}(3x^3+4x+4) is (9x^2+4)/(ln(3)(3x^3+4x+4))