Question

Differentiate from first principle:

$\left(vii\right)\sin (x+1)$

Answer 1

Given:

$f\left(x\right)=\sin (x+1)$

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

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{sin(x+h+1)-sin(x+1)}{h}$

Applying the formula,

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

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{2\cos \left(\frac{x+h+1+x+1}{2}\right)\sin \left(\frac{x+h+1-x-1}{2}\right)}{h}$

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

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

$\Rightarrow f^{\prime }\left(x\right)=\cos \left(\frac{2x+0+2}{2}\right)\times 1(\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1)$

$\Rightarrow f^{\prime }\left(x\right)=\cos (x+1)$

Therefore, the derivative of $\sin (x+1)$ is $\cos (x+1).$