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

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

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

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

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derivative of

f (x) = 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 )}

\frac{\partial}{\partial y} (5 x^{3} y^{6})

\frac{y ^{^{'}} + 0.3334 y}{( 20.004 x + 6000 )} = 20

2 x y + x^{3} = x \frac{d y}{d x}

f (x) = x^{3} - 3 (x)^{2} + 4 (x) + 5

(1 + x) cos (x), at x = 0

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