jika y=√2x+2 maka nilai y'(1)= tolong sama penjelasan dan caranya

y = √(2x+2)
y = (2x + 2)^(1/2)

Ingat ,
jika y = ( f(x) )^n
Maka
y' = n . ( f(x) )^(n-1) . ( f'(x))
y' = 1/2 . (2x + 2)^(1/2 - 1) . (2)
y' = (2x + 2)^( - 1/2)
y' = 1/ (2x+2)^1/2
y'
= [tex] \dfrac{1}{\sqrt{2x+2}} [/tex]