What is the derivative of y=log_{3}(8x-1) ?

The derivative of y=log_{3}(8x-1) is 8/(ln(3)(8x-1))

What is the first derivative of y=log_{3}(8x-1) ?

The first derivative of y=log_{3}(8x-1) is 8/(ln(3)(8x-1))

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derivative of

y = lo g_{3} (8 x - 1)

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

Solution steps

\frac{d}{d x} (lo g_{3} (8 x - 1))

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

\frac{1}{8 x - 1} \frac{d}{d x} (8 x - 1)

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

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

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

Simplify

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

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

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

y^{^{'}} + \frac{y}{6960} = 0.0025862069

\frac{d}{d x} (e^{(- x)})

\int 3 x d x

\frac{d}{d x} ((tan (x))^{s e c (x)})

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