Derivative of ln(1-x)
2026-02-28 06:10 Diff
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We can derive the derivative of ln(1-x) using proofs. To show this, we will use the basic rules of differentiation along with the chain rule

There are several methods we use to prove this, such as:

By First Principle Using Chain Rule We will now demonstrate that the differentiation of ln(1-x) results in -1/(1-x) using the above-mentioned methods:

By First Principle The derivative of ln(1-x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.

To find the derivative of ln(1-x) using the first principle, we will consider f(x) = ln(1-x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = ln(1-x), we write f(x + h) = ln(1-(x + h)).

Substituting these into equation (1), f'(x) = limₕ→₀ [ln(1-(x + h)) - ln(1-x)] / h = limₕ→₀ ln[(1-(x + h))/(1-x)] / h

Using the property of logarithms ln(a/b) = ln(a) - ln(b), f'(x) = limₕ→₀ ln[(1-x-h)/(1-x)] / h = limₕ→₀ ln[1 - h/(1-x)] / h

Using the approximation ln(1-u) ≈ -u for small u, f'(x) = limₕ→₀ [-h/(1-x)] / h = -1/(1-x)

Hence, proved. Using Chain Rule To prove the differentiation of ln(1-x) using the chain rule, Consider f(x) = 1-x, then ln(1-x) = ln(f(x)).

By the chain rule: d/dx [ln(f(x))] = (1/f(x)) * f'(x)

Let’s substitute f(x) = 1-x, then f'(x) = -1. d/dx (ln(1-x)) = (1/(1-x)) * (-1) = -1/(1-x)