# How do you differentiate #log(8x-1)#?

see below

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To differentiate log(8x-1), you can use the chain rule of differentiation. The derivative of log(u) with respect to x is (1/u) * du/dx.

So, applying the chain rule, the derivative of log(8x-1) with respect to x is:

d/dx [log(8x-1)] = (1/(8x-1)) * d/dx [8x-1]

Now, differentiate 8x-1 with respect to x:

d/dx [8x-1] = 8

Substitute this result back into the expression:

(1/(8x-1)) * 8 = 8/(8x-1)

Therefore, the derivative of log(8x-1) with respect to x is 8/(8x-1).

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To differentiate log(8x-1), we'll use the chain rule. The derivative of log(u) with respect to x is (1/u) * u', where u is a function of x and u' is the derivative of u with respect to x.

So, the differentiation of log(8x-1) with respect to x is:

d/dx [log(8x-1)] = (1/(8x-1)) * d/dx [8x-1]

Now, we'll find the derivative of 8x-1:

d/dx [8x-1] = 8

Substituting this back into our expression, we get:

d/dx [log(8x-1)] = (1/(8x-1)) * 8

Simplifying, we get:

d/dx [log(8x-1)] = 8 / (8x-1)

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