• 𝑑(𝑙𝑜𝑔_𝑎𝑥)/𝑑𝑥 = 1/(𝑥·𝑙𝑛(𝑎))

Proof

Let us assume that:

• 𝑓(𝑥) = 𝑙𝑜𝑔_𝑎𝑥

The derivative of a function 𝑓(𝑥) (which is denoted by 𝑓’(𝑥)) is given by the limit:

• 𝑓’(𝑥) = 𝑙𝑖𝑚_ℎ→0 [𝑓(𝑥 + ℎ) - 𝑓(𝑥)] / ℎ
• 𝑓’(𝑥) = 𝑙𝑖𝑚_ℎ→0 [𝑙𝑜𝑔_𝑎(𝑥 + ℎ) - 𝑙𝑜𝑔_𝑎(𝑥)] / ℎ
• 𝑓’(𝑥) = 𝑙𝑖𝑚_ℎ→0 [𝑙𝑜𝑔_𝑎[(𝑥 + ℎ)/𝑥]] / ℎ # Using a property of logarithms: logₐm - logₐn = logₐ(m/n)
• 𝑓’(𝑥) = 𝑙𝑖𝑚_ℎ→0 [𝑙𝑜𝑔_𝑎[1 + (ℎ/𝑥)]] / ℎ
• 𝑓’(𝑥) = 𝑙𝑖𝑚_𝑡→0 𝑙𝑜𝑔_𝑎(1 + 𝑡) / 𝑥𝑡 # ℎ = 𝑥𝑡 and ℎ→0, ℎ/𝑥→0 ⇒ 𝑡→0
• 𝑓’(𝑥) = 𝑙𝑖𝑚_𝑡→0 [1/(𝑥𝑡)]·𝑙𝑜𝑔_𝑎(1 + 𝑡)
• 𝑓’(𝑥) = 𝑙𝑖𝑚_𝑡→0 𝑙𝑜𝑔_𝑎[(1 + 𝑡)^1/(𝑥𝑡)] # using the property of logarithm, m·logₐa = logₐa^m
• 𝑓’(𝑥) = 𝑙𝑖𝑚_𝑡→0 𝑙𝑜𝑔_𝑎[[(1 + 𝑡)^1/𝑡]^1/𝑥] # using a property of exponents, a^mn = (a^m)ⁿ
• 𝑓’(𝑥) = 𝑙𝑖𝑚_𝑡→0 (1/𝑥)·𝑙𝑜𝑔_𝑎[(1 + 𝑡)^1/𝑡] # using the property of logarithm logₐa^m = m·logₐ a
• 𝑓’(𝑥) = (1/𝑥)·𝑙𝑖𝑚_𝑡→0 𝑙𝑜𝑔_𝑎[(1 + 𝑡)^1/𝑡] # the variable of the limit is 𝑡. So we can write (1/𝑥) outside of the limit
• 𝑓’(𝑥) = (1/𝑥)·𝑙𝑜𝑔_𝑎[𝑙𝑖𝑚_𝑡→0(1 + 𝑡)^1/𝑡]
• 𝑓’(𝑥) = (1/𝑥)·𝑙𝑜𝑔_𝑎(𝑒) # see Euler’s number
• 𝑓’(𝑥) = (1/𝑥)·[1/𝑙𝑜𝑔_𝑒(𝑎)]
• 𝑓’(𝑥) = 1/[𝑥·𝑙𝑜𝑔_𝑒𝑎]
• 𝑓’(𝑥) = 1/[𝑥·𝑙𝑛(𝑎)]