Derivative of ln(1/x)
2026-02-28 06:07 Diff
https://brightchamps.com/en-us/math/calculus/derivative-of-ln(1/x)

We can derive the derivative of ln(1/x) using proofs.

To show this, we will use the properties of logarithms along with the rules of differentiation.

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

By First Principle

Using Chain Rule

Using Logarithmic Differentiation

We will now demonstrate that the differentiation of ln(1/x) results in -1/x using the above-mentioned methods:

Using Chain Rule

To prove the differentiation of ln(1/x) using the chain rule, We use the formula: ln(1/x) = ln(x⁻¹) = -ln(x) Differentiating -ln(x), d/dx [-ln(x)] = -d/dx [ln(x)] d/dx [-ln(x)] = -1/x

Hence, proved.

Using Logarithmic Differentiation

To prove the differentiation of ln(1/x) using logarithmic differentiation,

Consider y = ln(1/x) Taking the derivative, dy/dx = d/dx [ln(x⁻¹)] dy/dx = d/dx [-ln(x)] dy/dx = -1/x

Hence, proved.

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(x + h) + ln(x)] / h

Using the property of logarithms,

ln(a) - ln(b) = ln(a/b), f'(x) = limₕ→₀ ln(x/(x + h)) / h

Converting to exponential form, this becomes, f'(x) = limₕ→₀ ln(1 - h/x) / h

Using the limit property, limₕ→₀ ln(1 - h/x) / h = -1/x f'(x) = -1/x

Hence, proved.