Name: Continuity Theorem
Uploaded: 2019-07-19
Duration: 5 min 49 s
Description: 5 is greater than 3, but 3 is not greater than 5. Similarly, if a function is differentiable, it is continuous, but the converse is not true. Watch this vide...

Question

Find the derivative of $\log (2 x + 1)$ w.r.t $x$ by using method of first principle at $x = 2$ .

Accepted Answer

Let $f (x) = \log (2 x + 1)$

We know that, the first principle of differentiation is $f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$ .

$∴ f^{'} (x) = \lim_{h \to 0} \frac{\log (2 (x + h) + 1) - \log (2 x + 1)}{h}$

$\Rightarrow f^{'} (x) = \lim_{h \to 0} \frac{\log (2 x + 2 h + 1) - \log (2 x + 1)}{h}$

$\Rightarrow f^{'} (x) = \lim_{h \to 0} \frac{\log (\frac{2 x + 2 h + 1}{2 x + 1})}{h} [∵ \log_{a} (\frac{x}{y}) = \log_{a} x - \log_{a} y]$

$\Rightarrow f^{'} (x) = \lim_{h \to 0} \frac{\log (\frac{2 x + 1 + 2 h}{2 x + 1})}{h}$

$\Rightarrow f^{'} (x) = \lim_{h \to 0} \frac{\log (1 + \frac{2 h}{2 x + 1})}{h}$

$\Rightarrow f^{'} (x) = \lim_{h \to 0} \frac{\log (1 + \frac{2 h}{2 x + 1})}{\frac{2 h}{2 x + 1} \times \frac{2 x + 1}{2}}$

$\Rightarrow f^{'} (x) = \lim_{h \to 0} \frac{\log (1 + \frac{2 h}{2 x + 1})}{\frac{2 h}{2 x + 1}} \times \frac{2}{2 x + 1}$

$\Rightarrow f^{'} (x) = \frac{2}{2 x + 1} \lim_{h \to 0} \frac{\log (1 + \frac{2 h}{2 x + 1})}{\frac{2 h}{2 x + 1}} [∵ \lim_{x \to 0} (a . f (x)) = a \lim_{x \to 0} f (x)]$

$\Rightarrow f^{'} (x) = \frac{2}{2 x + 1} \times 1 [∵ \lim_{x \to 0} \frac{\log (1 + x)}{x} = 1]$

$\Rightarrow f^{'} (x) = \frac{2}{2 x + 1}$

Now putting $x = 2$ , we get

$\Rightarrow f^{'} (2) = \frac{2}{5}$

Hence, the required value is $\frac{2}{5}$ .

Find the derivative of $\log (2 x + 1)$ w.r.t $x$ by using method of first principle at $x = 2$ .

Important Questions on Diffrentiation

Learn and Prepare for any exam you want !

Find the derivative of log(2x+1) w.r.t x by using method of first principle at x=2.

Important Questions on Diffrentiation

Learn and Prepare for any exam you want !

Find the derivative of $\log (2 x + 1)$ w.r.t $x$ by using method of first principle at $x = 2$ .