Question

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

Answer 1

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

Explanation:

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)`.