Question

Differentiate from the first principle the derivative of $\ln \left( x \right)?$

Answer 1

Here we have to find the derivative of a function using the first principle of differentiation. According to the first principle of differentiation, if we differentiate a function $f\left( x \right)$ so the differentiation ${f'}\left( x \right)$ represents the slope of a tangent on a curve. The formula is given as: ${f'}\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$ . So, use this concept to get the solution of the given problem.

Complete step by step answer:
The given function is $\ln \left( x \right)$
So let $f\left( x \right) = \ln \left( x \right){\text{ }} - - - \left( 1 \right)$
Now we have to find the derivative of $\ln \left( x \right)$ using the first principle of differentiation.
According to the first principle of differentiation, the derivative of a function can be evaluated using the formulas as:
${f'}\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$
So, on differentiating equation $\left( 1 \right)$ using the formula of first principle we get
$\dfrac{d}{{dx}}\ln \left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - \ln \left( x \right)}}{h}{\text{ }} - - - \left( 2 \right)$
For finding $f\left( {x + h} \right)$ replace $x$ by $x + h$ in the given function
Therefore, we get
$f\left( {x + h} \right) = \ln \left( {x + h} \right)$
On substituting in the equation $\left( 2 \right)$ we get
$\dfrac{d}{{dx}}\ln \left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\ln \left( {x + h} \right) - \ln \left( x \right)}}{h}{\text{ }} - - - \left( 3 \right)$
Now we know that
${\log _a}\left( {\dfrac{x}{y}} \right) = {\log _a}x - {\log _a}y$
Therefore, from equation $\left( 3 \right)$ we have
$\dfrac{d}{{dx}}\ln \left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\ln \left( {\dfrac{{x + h}}{x}} \right)}}{h}$
$\Rightarrow \dfrac{d}{{dx}}\ln \left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\ln \left( {\dfrac{x}{x} + \dfrac{h}{x}} \right)}}{h}$
Now multiply and divide by $x$ in the denominator, we have
$\Rightarrow \dfrac{d}{{dx}}\ln \left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\ln \left( {\dfrac{x}{x} + \dfrac{h}{x}} \right)}}{{h \cdot \dfrac{x}{x}}}$
$\Rightarrow \dfrac{d}{{dx}}\ln \left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\ln \left( {\dfrac{x}{x} + \dfrac{h}{x}} \right)}}{{\dfrac{h}{x} \cdot x}}$
Solving numerator term, we get
$\Rightarrow \dfrac{d}{{dx}}\ln \left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\ln \left( {1 + \dfrac{h}{x}} \right)}}{{\dfrac{h}{x} \cdot x}}$
Now we know that
$\mathop {\lim }\limits_{h \to 0} \dfrac{{{{\log }_e}\left( {1 + \dfrac{h}{x}} \right)}}{{\dfrac{h}{x}}} = 1$
Therefore, from the above equation, we get
$\Rightarrow \dfrac{d}{{dx}}\ln \left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{x}$
Now $\dfrac{1}{x}$ doesn’t matter in this limit, so we can bring $\dfrac{1}{x}$ entirely out of the limit.
Therefore, we get
$\Rightarrow \dfrac{d}{{dx}}\ln \left( x \right) = \dfrac{1}{x}\mathop {\lim }\limits_{h \to 0} 1$
$\Rightarrow \dfrac{d}{{dx}}\ln \left( x \right) = \dfrac{1}{x}$
Therefore, the derivative of $\ln \left( x \right)$ is $\dfrac{1}{x}$

Note:
Differentiation using first principle is also known as the delta method. The first principle is nothing, but it is the first derivative of the function. The first principle of differentiation helps us to evaluate the derivative of a function using limits. Also note in the formula, in place of $h$ there is no need to substitute anything. We need to just substitute $x + h$ in place of $x$ Also at last always remember if the value of the function has $h$ then substitute the limit value to that.