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

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

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

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

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

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

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\frac{d ^{2} y}{d x ^{2}} - 9 \frac{d y}{d x} + 4 y = x e^{x}

sin (x) + \frac{3}{5} cot (x)

\frac{1}{s + 6}

\frac{\partial}{\partial y} (\frac{y}{x + y + z})

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