Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

Find the derivation of each of the following from the first principle:

Question:

Find the derivation of each of the following from the first principle:

$\sin (2 x+3)$

Solution:

Let $f(x)=\sin (2 x+3)$

We need to find the derivative of $f(x)$ i.e. $f^{\prime}(x)$

We know that,

$\mathrm{f}^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}(\mathrm{x})}{\mathrm{h}}$ …(i)

$f(x)=\sin (2 x+3)$

$f(x+h)=\sin [2(x+h)+3]$

Putting values in (i), we get

$\mathrm{f}^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{\sin [2(\mathrm{x}+\mathrm{h})+3]-\sin (2 \mathrm{x}+3)}{\mathrm{h}}$

Using the formula:

$\sin A-\sin B=2 \sin \frac{A-B}{2} \cos \frac{A+B}{2}$

$=\lim _{h \rightarrow 0} \frac{2 \sin \frac{2(x+h)+3-(2 x+3)}{2} \cos \frac{2(x+h)+3+2 x+3}{2}}{h}$

$=\lim _{h \rightarrow 0} \frac{2 \sin \frac{2 x+2 h+3-2 x-3}{2} \cos \frac{2 x+2 h+6+2 x}{2}}{h}$

$=\lim _{h \rightarrow 0} \frac{2 \sin \frac{2 h}{2} \cos \frac{4 x+2 h+6}{2}}{h}$

$=\lim _{h \rightarrow 0} \frac{2 \sin (h) \cos (2 x+h+3)}{h}$

$=2 \lim _{h \rightarrow 0} \frac{\sin h}{h} \times \lim _{h \rightarrow 0} \cos (2 x+h+3)$

$=2(1) \times \lim _{h \rightarrow 0} \cos (2 x+h+3)$

$\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

Putting h = 0, we get

$=2 \cos (2 x+0+3)$

$=2 \cos (2 x+3)$

Hence, f’(x) = 2cos (2x + 3)

 

Leave a comment

None
Free Study Material