Derivative of log x Proof by First Principle
We will prove that d/dx(logₐ x) = 1/(x ln a) using the first principle (definition of the derivative).
Proof:
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)Thus, we proved that the derivative of logₐ x is 1 / (x ln a) by the first principle.
