Compute f '(a) algebraically for the given value of a.

f(x) = 2x^2 + x; a = 9

thanks, 10pointss!

f'(a)=lim {(f(a+h)-f(a))/h} as h tends to 0
f'(a)=lim { [2(9+h)^2+9+h-(2*9^2+9)]/h} as h tends to 0
f'(9)=lim { [2(81+18h+h^2)+9+h-2*81-9]/h} as h tends to 0
f'(9)=lim { [37h+h^2]/h} as h tends to 0
f'(9)=lim {37+h} as h tends to 0
f'(9)=37

f'(x)=4x+1
f'(9)=36+1=37