What is the derivative of $log(8x-1)$ ?

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Answer 1 · 2015-07-30T21:07:30

$8/(ln(10)(8x-1)) approx 3.474/(8x-1)$

Assuming $log(x)=log_{10}(x)$ , $d/dx(log(x))=1/(ln(10)x)$ , where $ln(10)=log_{e}(10)$ (and $e approx 2.71828$ ).

By the Chain Rule, $d/dx(f(g(x)))=f'(g(x))*g'(x)$ , we get

$d/dx(log(8x-1))=1/(ln(10)(8x-1)) * d/dx(8x-1)$

$=8/(ln(10)(8x-1)) approx 3.474/(8x-1)$ .

What is the derivative of log(8x-1)log(8x-1)?