Proof of Derivative of log x
The derivative of log x can be proved using the following 3 ways:
- By First Principle of Derivative
- By Implicit Differentiation Method
- By ln x Formula
Derivative of log x by First Principle of Derivative
To prove derivative of log x using First Principle of Derivative, we will use basic limits and log formulas which are listed below:
- logₐ m – logₐ n = logₐ (m/n)
- m logₐ a = logₐ am
- amn = (am)n
- logₐ am = m logₐ a
- limt→0 [(1 + t)1/t] = e
Let’s start the proof for the derivative of log x , assume f(x)=logax
By First Principle of Derivative
f'(x) = limh→0 [f(x + h) – f(x)] / h
Since 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] / h
⇒ f'(x) = limh→0 [logₐ [(x + h) / x] ] / h { By 1 }
⇒ limh→0 [logₐ (1 + (h/x))] / h
Assume 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)
⇒ f'(x) = limt→0 logₐ (1 + t)1/(xt) { By 2 }
⇒ f'(x) = limt→0 logₐ [(1 + t)1/t]1/x { By 3 }
⇒ f'(x) = limt→0 (1/x) logₐ [(1 + t)1/t] { By 4 }
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,. Therefore,
f'(x) = (1/x) logₐ e { By 5 }
⇒ (1/x) (1/logₑ a) (because ‘a’ and ‘e’ are interchanged)
⇒ (1/x) (1/ ln a) (because logₑ = ln)
⇒ 1 / (x ln a)
Therefore for f(x)=log x , f'(x) =1/(x ln 10)
Derivative of log x by Implicit Differentiation Method
The derivative of log x can be proved using the implicit differentiation method. In this method, if we are given an implicit function, then we take the derivative on both sides of the equation with respect to the independent variable. We will use basic logarithmic formulas which are listed below:
- dy/dx(ay) =ay ln a
- y = logax ⇒ ay=x
Let’s start the proof for the derivative of log x , assume f(x)=y=logax
By Implicit Differentiation Method
f(x)=y=logax
⇒ ay=x
Taking derivative on both sides with respect to “x”
⇒ d/dx (ay) = d/dx (x)
By using the chain rule,
(ay ln a) dy/dx = 1 { By 1 }
⇒ dy/dx = 1/(ay ln a)
But we have ay = x. Therefore,
dy/dx = 1 / (x ln a)
Therefore for f(x)=log x , f'(x) = 1 / (x ln 10)
Derivative of log x by ln x Formula
We know that the derivative of ln x is 1/x. We can convert log into ln using change of base rule. We will be using basic log formulas which are listed below:
- log a x = (logₑ x) / (logₑ a)
- logₑ = ln
Let’s start the proof for the derivative of log x , assume f(x)=logax
By change of base rule
f(x) = (logₑ x) / (logₑ a) { By 1 }
⇒ f(x) = (ln x) / (ln a) { By 2 }
Now we will find its derivative.
f'(x) = d/dx [(ln x) / (ln a)]
⇒ 1/ (ln a) d/dx (ln x)
⇒ 1 / (ln a) · (1/x)
⇒ 1 / (x ln a)
Therefore for f(x)= log x , f'(x) =1 / (x ln 10)
Derivative of Log x: Formula and Proof
Derivative of log x is 1/x. Log x Derivative refers to the process of finding change in log x function to the independent variable. The specific process of finding the derivative for log x functions is referred to as logarithmic differentiation. The function log x typically refers to the natural logarithm of x, which is the logarithm to the base e, where e is Euler’s number, approximately equal to 2.71828.
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