# Find the derivative of the given function from first principle : $\sin (x+1)$

Let $f(x) = \sin (x+1)$. Accordingly, $f(x+h)= \sin (x+h+1)$
By first principle,
$f'(x) = \lim\limits_{h \to 0} \large\frac{f(x+h)-f(x)}{h}$
$= \lim\limits_{h \to 0} \large\frac{1}{h}$$[\sin (x+h+1)- \sin (x+1)] = \lim\limits_{h \to 0} \large\frac{1}{h}$$ \bigg[ 2 \cos \bigg( \large\frac{x+h+1+x+1}{2} \bigg) $$\sin \bigg( \large\frac{x+h+1-x-1}{2} \bigg) \bigg] = \lim\limits_{h \to 0} \large\frac{1}{h}$$ \bigg[ 2 \cos \bigg( \large\frac{2x+h+2}{2} \bigg) $$\sin \bigg( \large\frac{h}{2} \bigg) \bigg] = \lim\limits_{h \to 0} \bigg[ \cos \bigg( \large\frac{2x+h+2}{2} \bigg)$$. \large\frac{sin\bigg( \large\frac{h}{2}\bigg)}{\bigg(\large\frac{h}{2}\bigg)} \bigg]$
$= \lim\limits_{h \to 0} \cos \bigg( \large\frac{2x+h+2}{2} \bigg) $$.\lim\limits_{ \large\frac{h}{2} \to 0} \large\frac{sin\bigg( \large\frac{h}{2}\bigg)}{\bigg(\large\frac{h}{2}\bigg)}$$\qquad \qquad \bigg[ As \: h \to 0 \Rightarrow \large\frac{h}{2}$$\to 0 \bigg] = \cos \bigg( \large\frac{2x+0+2}{2} \bigg)$$.1 \qquad \qquad \bigg[ \lim\limits_{x \to 0} \large\frac{ \sin x}{x}$$=1 \bigg]$
$= \cos (x+1)$