Question

Find the derivative by first principle:
$\rm \sqrt{\sin x}$

Answer 1

HOPE IT HELPS U ✌️✌️

Answer 2

Answer:

f'(x)  =  Cosx / 2√Sinx

Step-by-step explanation:

f(x)  = √Sinx

f'(x)  = Lim h-0   (f(x + h)  - f(x) ) /h

=> f'(x)  = Lim h-0    (√Sin(x + h)  - √Sinx )/h

Mu;tiplying & Divding by √Sin(x + h)  + √Sinx

=> f'(x)  = Lim h-0    (√Sin(x + h)  - √Sinx )/h  * (√Sin(x + h)  + √Sinx)/((√Sin(x + h)  + √Sinx)

=> f'(x)  = Lim h-0   (Sin(x + h) - Sinx) / (h (√Sin(x + h)  + √Sinx))

Using Sin (A - B) = SinACosB  + CosASinB

=> f'(x)  = Lim h-0   (Sin(x + h) - Sinx) / (h (√Sin(x + h)  + √Sinx))

=> f'(x)  =  Lim h-0   (SinxCosh + CosxSinh - Sinx ) / (h (√Sin(x + h)  + √Sinx))

=> f'(x)  =  Lim h-0   (Sinx(Cosh - 1)/h + CosxSinh/h ) / ( (√Sin(x + h)  + √Sinx))

putting (Cosh - 1)/h = 0  & Sinh/h =  1 & h = 0

=> f'(x)  =  Cosx / 2√Sinx