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

1 Answer

Jul 30, 2015

#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)#.

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