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

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

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

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

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

\frac{2 x}{ln ( 6 ) ( x ^{2} + 2 )}

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int (\frac{10 x}{1 + 5 x ^{2}}) d x

\frac{2 x ^{3}}{3}

\int_{1}^{3} (x^{2} + 2 x - 5) d x

\frac{d}{d x} (\frac{e ^{x}}{3 x ^{2}})

x \to 0 lim (6 x^{2} (cot (3 x)) (csc (2 x)))

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