Rivalry2

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Original 2026-01-01

Modified 2026-02-28

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

2 To show this, we will use the properties of<a>logarithms</a>along with the rules of differentiation.

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

4 By First Principle

5 Using Chain Rule

6 Using Logarithmic Differentiation

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

8 Using Chain Rule

9 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

10 Hence, proved.

11 Using Logarithmic Differentiation

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

13 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

14 Hence, proved.

15 By First Principle

16 The derivative of ln(1/x) can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.

17 To find the derivative of ln(1/x) using the first principle, we will consider f(x) = ln(1/x).

18 Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)

19 Given that f(x) = ln(1/x), we write f(x + h) = ln(1/(x + h)).

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

21 Using the property of logarithms,

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

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

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

25 Hence, proved.