What is the derivative of log_{6}(3x^3+4x^2+4x) ?

The derivative of log_{6}(3x^3+4x^2+4x) is (9x^2+8x+4)/(ln(6)(3x^3+4x^2+4x))

What is the first derivative of log_{6}(3x^3+4x^2+4x) ?

The first derivative of log_{6}(3x^3+4x^2+4x) is (9x^2+8x+4)/(ln(6)(3x^3+4x^2+4x))

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\frac{d}{d x} (lo g_{6} (3 x^{3} + 4 x^{2} + 4 x))

\frac{9 x ^{2} + 8 x + 4}{ln ( 6 ) ( 3 x ^{3} + 4 x ^{2} + 4 x )}

Solution steps

\frac{d}{d x} (lo g_{6} (3 x^{3} + 4 x^{2} + 4 x))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 3 x ^{3} + 4 x ^{2} + 4 x )}{ln ( 6 )})

Take the constant out:

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

= \frac{1}{ln ( 6 )} \frac{d}{d x} (ln (3 x^{3} + 4 x^{2} + 4 x))

Apply the chain rule:

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

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

\frac{d}{d x} (3 x^{3} + 4 x^{2} + 4 x) = 9 x^{2} + 8 x + 4

= \frac{1}{ln ( 6 )} \cdot \frac{1}{3 x ^{3} + 4 x ^{2} + 4 x} (9 x^{2} + 8 x + 4)

Simplify

\frac{1}{ln ( 6 )} \cdot \frac{1}{3 x ^{3} + 4 x ^{2} + 4 x} (9 x^{2} + 8 x + 4) : \frac{9 x ^{2} + 8 x + 4}{ln ( 6 ) ( 3 x ^{3} + 4 x ^{2} + 4 x )}

= \frac{9 x ^{2} + 8 x + 4}{ln ( 6 ) ( 3 x ^{3} + 4 x ^{2} + 4 x )}

\int 64 - 5 x^{2} d x

\int \frac{( x ^{2} )}{20 - 4 x + x ^{2}} d x

\int \frac{ln ( 16 x )}{x} d x

y^{^{'}} = \frac{e ^{t}}{2 y}

\int (sin^{4} (x) cos (x)) d x

What is the derivative of log_{6}(3x^3+4x^2+4x) ?
The derivative of log_{6}(3x^3+4x^2+4x) is (9x^2+8x+4)/(ln(6)(3x^3+4x^2+4x))
What is the first derivative of log_{6}(3x^3+4x^2+4x) ?
The first derivative of log_{6}(3x^3+4x^2+4x) is (9x^2+8x+4)/(ln(6)(3x^3+4x^2+4x))