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

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

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

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

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derivative of

lo g_{3} (2 x^{2} + 4 x + 3)

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

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

n = 1 \sum \infty \frac{9 9 ^{n} ( n ^{5} + 2 )}{n ^{2} 1 0 ^{2 n}}

f (x) = \frac{11}{ln ( x )}

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

\frac{d}{d x} ((6 x + 7) (7 x - 8))

\int x^{2} x^{e} d x

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