What is the derivative of log_{6}(cot(8x)) ?

The derivative of log_{6}(cot(8x)) is -(8csc^2(8x))/(ln(6)cot(8x))

What is the first derivative of log_{6}(cot(8x)) ?

The first derivative of log_{6}(cot(8x)) is -(8csc^2(8x))/(ln(6)cot(8x))

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\frac{d}{d x} (lo g_{6} (cot (8 x)))

- \frac{8 csc ^{2} ( 8 x )}{ln ( 6 ) cot ( 8 x )}

Solution steps

\frac{d}{d x} (lo g_{6} (cot (8 x)))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( cot ( 8 x ) )}{ln ( 6 )})

Take the constant out:

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

= \frac{1}{ln ( 6 )} \frac{d}{d x} (ln (cot (8 x)))

Apply the chain rule:

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

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

\frac{d}{d x} (cot (8 x)) = - csc^{2} (8 x) \cdot 8

= \frac{1}{ln ( 6 )} \cdot \frac{1}{cot ( 8 x )} (- csc^{2} (8 x) \cdot 8)

Simplify

\frac{1}{ln ( 6 )} \cdot \frac{1}{cot ( 8 x )} (- csc^{2} (8 x) \cdot 8) : - \frac{8 csc ^{2} ( 8 x )}{ln ( 6 ) cot ( 8 x )}

= - \frac{8 csc ^{2} ( 8 x )}{ln ( 6 ) cot ( 8 x )}

What is the derivative of log_{6}(cot(8x)) ?
The derivative of log_{6}(cot(8x)) is -(8csc^2(8x))/(ln(6)cot(8x))
What is the first derivative of log_{6}(cot(8x)) ?
The first derivative of log_{6}(cot(8x)) is -(8csc^2(8x))/(ln(6)cot(8x))