Question

Find the derivative of the following functions from first
principles.
sin (x+1)​

Answer 1

Answer:

$\sf f' ( x ) = \cos\left(x+1\right)$

Step-by-step explanation:

Given :

f ( x ) = sin ( x + 1 )

We have first principle :

if f ( x ) = y

$\sf f' ( x ) = \lim_{h \to o} \dfrac{f ( x + h ) - f ( x )}{h}$

We have :

f ( x ) = sin ( x + 1 )

f ( x + h ) = sin ( x + h + 1 )

$\sf f' ( x ) = \lim_{h \to o} \dfrac{\sin(x+h+1) - \sin(x+1)}{h}$

Using sum to product formula :

$\sf \sin \ C-\sin \ D= 2\cos \ \left(\dfrac{C+D}{2}\right).\sin\left(\dfrac{C-D}{2}\right)$

$\sf f' ( x ) = \lim_{h \to o} \ \left(\dfrac{2 \cos \ \left(\dfrac{x+h+1+x+1}{2} \right).\sin\left(\dfrac{x+1+h-x-1}{2} \right)}{h} \right)$

$\sf f' ( x ) = \lim_{h \to o} \ \left(\dfrac{2 \cos \ \left(\dfrac{2x+2+h}{2} \right).\sin\left(\dfrac{h}{2} \right)}{h} \right)$

Multiply and divide by h / 2. Also separate the limit.

We know :

$\sf \lim_{x \to 0} \ \dfrac{\sin x}{x} = 1$

$\sf f' ( x ) = \lim_{h \to o} \ \cos\dfrac{2x+2+h}{2}$

Putting h = 0

$\sf f' ( x ) = \cos\dfrac{2x+2+0}{2}$

$\sf f' ( x ) = \cos\left(x+1\right)$

Hence we get required answer.