Question

Differentiate from first principle:

$\left(xi\right)\sin (2x-3)$

Answer 1

Given:

$f\left(x\right)=\sin (2x-3)$

The derivative of a function $f\left(x\right)$ is defined as:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$


Putting $f\left(x\right)$ in the above expression, we get:

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\sin (2x+2h-3)-\sin (2x-3)}{h}$

Applying formula

$\sin C-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right),$ we get:

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{2\cos \left(\frac{2x+2h-3+2x-3}{2}\right)\sin \left(\frac{2x+2h-3-2x+3}{2}\right)}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}=\frac{2\cos \left(\frac{4x+2h-6}{2}\right)\sin \left(h\right)}{h}$

$\Rightarrow f^{\prime }\left(x\right)=2\cos \left(\frac{4x+0-6}{2}\right)\left(1\right)$

$(\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1)$

$\Rightarrow f^{\prime }\left(x\right)=2\cos (2x-3)$