jika f(x) = [tex]4x \sqrt{ {x}^{2} + 16 } [/tex] Maka f'(3) =? Tolong dijawab terimakasih....

u = 4x ---> u' = 4
v = √(x² + 16) ---> v' = (1/2)(x² + 16)^(-1/2).(2x)
v' = x/√(x² + 16)
f'(x) = u'v + v'u = 4(4x)√(x² + 16) + (4x²/√(x² + 16))
= 16x√(x² + 16) + 4x²/√(x² + 16)
f'(3) = 16(3)/√(3² + 16) + 4(3²)/√(3² + 16) = 84/5