Question

Differentiate using first principle method.

f(x) = ln(x)/x​

Answer 1

Answer:

$\sf\dfrac{1-\ln(x)}{x^2}$

Step-by-step explanation:

Function we are provided to differentiate,

$\sf f(x) = \dfrac{\ln(x)}{x}$

By the definition of first principles, we know that:

$\boxed{\tt f'(x) = \lim\limits_{h\to0} \dfrac{f(x+h) - f(x)}{h}}$

By using this we get:

${\sf \implies f'(x) = \lim\limits_{h\to0} \dfrac{\frac{\ln(x+h)}{(x+h)} - \frac{\ln(x)}{x}}{h}}$

${\sf \implies f'(x) = \lim\limits_{h\to0} \dfrac{\frac{x\ln(x+h) - (x+h)\ln(x)}{x(x+h)}}{h}}$

${\sf \implies f'(x) = \lim\limits_{h\to0} \dfrac{\frac{x\ln(x+h) - x\ln(x) - h\ln(x)}{x(x+h)}}{h}}$

${\sf \implies f'(x) = \lim\limits_{h\to0}\dfrac{x[\ln(x+h) - \ln(x)] - h\ln(x)}{h\cdot x(x+h)}}$

${\sf \implies f'(x) = \lim\limits_{h\to0}\dfrac{x[\ln(x+h) - \ln(x)]}{h\cdot x(x+h)} -\dfrac{ h\ln(x)}{h\cdot x(x+h)}}$

${\sf \implies f'(x) = \lim\limits_{h\to0}\dfrac{\ln(x+h) - \ln(x)}{h(x+h)} -\dfrac{ \ln(x)}{x(x+h)}}$

Now, we will use the property of logarithm,

$\boxed{\tt \log(a) - \log(b) = \log\frac{a}{b}}$

Using this, we get:

${\sf \implies f'(x) = \lim\limits_{h\to0}\dfrac{\ln(\frac{x+h}{x}) }{h(x+h)} -\dfrac{ \ln(x)}{x(x+h)}}$

${\sf \implies f'(x) = \lim\limits_{h\to0}\dfrac{\ln(1 + \frac{h}{x}) }{h(x+h)} -\lim\limits_{h\to0}\dfrac{ \ln(x)}{x(x+h)}}$

${\sf \implies f'(x) = \lim\limits_{h\to0}\dfrac{\ln(1 + \frac{h}{x}) }{x \cdot \frac{h}{x} (x+h)} -\dfrac{ \ln(x)}{ {x}^{2} }}$

${\sf \implies f'(x) = \lim\limits_{h\to0}\dfrac{\ln(1 + \frac{h}{x}) }{\frac{h}{x}} \cdot \dfrac{1}{x(x + h)} -\dfrac{ \ln(x)}{ {x}^{2} }}$

Now apply special limit:

$\boxed{\tt\lim_{x\to0} \dfrac{\ln(1+x)}{x} = 1}$

Using this, we get:

${\sf \implies f'(x) = \lim\limits_{h\to0} \dfrac{1}{x(x + h)} -\dfrac{ \ln(x)}{ {x}^{2} }}$

${\sf \implies f'(x) = \dfrac{1}{ {x}^{2} } -\dfrac{ \ln(x)}{ {x}^{2} }}$

$\underline{ \underline{\sf \implies f'(x) = \dfrac{1 - \ln(x)}{ {x}^{2} }}}$

This is the required derivative.