Given:
f(x)=√sin(3x+1)
The derivative of a function f(x) is defined as:
f′(x)=limh→0f(x+h)−f(x)h
Putting f(x) in above expression, we get:
f′(x)=limh→0√sin(3(x+h)+1)−√sin(3x+1)h
Rationalizing the numerator, we get:
f′(x)=limh→0√sin(3(x+h)+1)−√sin(3x+1)h×√sin(3x+3h+1)+√sin(3x+1)√sin(3x+3h+1)+√sin(3x+1)
⇒f′(x)=limh→0sin(3x+3h+1)−sin(3x+1)h(√sin(3x+3h+1)+√sin(3x+1))
Applying the formula,
sinc−sind=2cos(c+d2)sin(c−d2), we get
⇒f′(x)=limh→02cos(3x+3h+1+3x+12)sin(3x+3h+1−3x−12)h√(sin(3x+3h+1)+√sin(3x+1))
⇒f′(x)=limh→02cos(6x+3h+22)sin3h2h(√sin(3x+3h+1)+√sin(3x+1))
⇒f′(x)=limh→02cos(6x+3h+22)(√sin(3x+3h+1)+√sin(3x+1)).sin3h23h2.32
⇒f′(x)=3cos(6x+22)(√sin(3x+1)+√sin(3x+1))
[∵limh→0sinhh=1]
⇒f′(x)=3cos(3x+1)2√sin(3x+1)