Let us assume that f(x) = logₐ x.
 By first principle, the derivative of a function f(x) (which is denoted by f'(x)) is given by the limit,
 f'(x) = limₕ→₀ [f(x + h) - f(x)] / hSince f(x) = logₐ x, we have f(x + h) = logₐ (x + h).
Substituting these values in the equation of first principle,
 f'(x) = limₕ→₀ [logₐ (x + h) - logₐ x] / hUsing a property of logarithms, logₐ m - logₐ n = logₐ (m/n). By applying this,
f'(x) = limₕ→₀ [logₐ [(x + h) / x] ] / h
       = lim ₕ→₀ [logₐ (1 + (h/x))] / hAssume that h/x = t. From this, h = xt.
 When h→0, h/x→0 ⇒ t→0.Then the above limit becomes
f'(x) = limₜ→₀ [logₐ (1 + t)] / (xt)
       = limₜ→₀ 1/(xt) logₐ (1 + t)By using property of logarithm, m logₐ a = logₐ am. By applying this,
f'(x) = limₜ→₀ logₐ (1 + t)1/(xt)
By using a property of exponents, amn = (am)n. By applying this,
f'(x) = limₜ→₀ logₐ [(1 + t)1/t]1/x
By applying the property logₐ am = m logₐ a,
f'(x) = limₜ→₀ (1/x) logₐ [(1 + t)1/t]
Here, the variable of the limit is 't'. So we can write (1/x) outside of the limit.
f'(x) = (1/x) limₜ→₀ logₐ [(1 + t)1/t] = (1/x) logₐ limₜ→₀ [(1 + t)1/t]
Using one of the formulas of limits, limₜ→₀ [(1 + t)1/t] = e. Therefore,
f'(x) = (1/x) logₐ e
       = (1/x) (1/logₑ a) (because 'a' and 'e' are interchanged)
       = (1/x) (1/ ln a) (because logₑ = ln)
       = 1 / (x ln a)