What is the derivative of log_{8}(x^2-3x) ?

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

What is the first derivative of log_{8}(x^2-3x) ?

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

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derivative of

lo g_{8} (x^{2} - 3 x)

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

Solution steps

\frac{d}{d x} (lo g_{8} (x^{2} - 3 x))

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int (ln (x + 1)) d x

\int sin (y) cos (y) d y

\frac{d y}{d x}, x = e^{y}

\int \frac{5}{4 + 9 t ^{2}} d t

y^{3} - (18 x + 6) + 3 x y^{2} y^{^{'}} = 0

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