Question

find the derivative of the function.
\\( y = log_8(6x + 1) \\)
\\( y = \\)
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find the derivative of the function.
\\( y = log_2(x^2 - 9x) \\)
\\( y = \\)
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Explanation:

Step1: Apply log derivative rule

For $y=\log_b(u)$, $y'=\frac{1}{u\ln b}\cdot u'$. For $y=\log_8(6x+1)$, $u=6x+1$, $b=8$.

Step2: Compute $u'$

$u'=\frac{d}{dx}(6x+1)=6$

Step3: Substitute into formula

$y'=\frac{1}{(6x+1)\ln 8}\cdot6=\frac{6}{(6x+1)\ln 8}$

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Step1: Apply log derivative rule

For $y=\log_2(x^2-9x)$, $u=x^2-9x$, $b=2$. Use $y'=\frac{1}{u\ln b}\cdot u'$.

Step2: Compute $u'$

$u'=\frac{d}{dx}(x^2-9x)=2x-9$

Step3: Substitute into formula

$y'=\frac{1}{(x^2-9x)\ln 2}\cdot(2x-9)=\frac{2x-9}{(x^2-9x)\ln 2}$

Answer:

1. $\frac{6}{(6x+1)\ln 8}$
2. $\frac{2x-9}{(x^2-9x)\ln 2}$