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

The derivative of log_{3}(x-4x^2) is (1-8x)/(ln(3)(x-4x^2))

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

The first derivative of log_{3}(x-4x^2) is (1-8x)/(ln(3)(x-4x^2))

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derivative of

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

\frac{1 - 8 x}{ln ( 3 ) ( x - 4 x ^{2} )}

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (x - 4 x^{2}) = 1 - 8 x

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

Simplify

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

= \frac{1 - 8 x}{ln ( 3 ) ( x - 4 x ^{2} )}

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

x \to 0 - lim (x ln (x))

\int_{3}^{8} (x^{3} - π x^{2}) d x

n = 0 \sum \infty 9 ((\frac{5}{12})^{n})

\int \frac{x ^{6} e ^{x}}{x} d x

What is the derivative of log_{3}(x-4x^2) ?
The derivative of log_{3}(x-4x^2) is (1-8x)/(ln(3)(x-4x^2))
What is the first derivative of log_{3}(x-4x^2) ?
The first derivative of log_{3}(x-4x^2) is (1-8x)/(ln(3)(x-4x^2))