Calculus Derivatives Question

f(t)=√(4t + 8)

Find f ''' (1/4) <--- Find the 3rd derivative of f(1/4)?

I'd implicitly derive

f(t) = (4t + 8)^(1/2)
f(t)^2 = 4t + 8
2 * f(t) * f'(t) = 4
f(t) * f'(t) = 2
f'(t) = 2 / f(t)
f''(t) = (f(t) * 0 - 2 * f'(t)) / f(t)^2
f''(t) = -2 * f'(t) / f(t)^2
f''(t) = -2 * (2/f(t)) / (f(t)^2)
f''(t) = -4 / f(t)^3

f'''(t) = (f(t)^3 * 0 + 4 * 3 * f(t)^2 * f'(t)) / f(t)^6
f'''(t) = 12 * f'(t) / f(t)^4
f'''(t) = 12 * (2/f(t)) / f(t)^4
f'''(t) = 24 / f(t)^5
f'''(t) = 24 / (4t + 8)^(5/2)
f'''(1/4) = 24 / (4 * (1/4) + 8)^(5/2)
f'''(1/4) = 24 / (1 + 8)^(5/2)
f'''(1/4) = 24 / 9^(5/2)
f'''(1/4) = 24 / 243

f(t)=√(4t+8)=(4t+8)^(1/2)
f'(t)=4(1/2)(4t+8)^(-1/2)
f''(t)=4*4*(1/2)*(-1/2)(4t+8)^(-3/2)
f'''(t)=4*4*4*(1/2)*(-1/2)*(-3/2)(4t+8…

f'''(1/4)=24(4(1/4)+8)^(-5/2)
f'''(1/4)=24(9)^(-5/2)
f'''(1/4)=24*(1/243)
f'''(1/4)=24/243
f'''(1/4)=8/81 which is approx .0988

f(t)=√(4t + 8) = 2√(t + 2) = 2(t + 2)^(1/2)
f’(t) = 2(1/2)(t + 2)^(-1/2) = (t + 2)^(-1/2)
f’’(t) = (-1/2) (t + 2)^(-3/2)
f”’(t) = (-1/2)(-3/2) (t + 2)^(-5/2) = ¾(t + 2)^(-5/2)

f”’(¼) = ¾(¼ + 2)^(-5/2) = ¾[√(9/4)]⁻⁵ = ¾[3/2]⁻⁵

= ¾[32/243] = 24/243

This is a stick-to-it and keep good track of the coefficients exercise.

f(t)=√(4t + 8)
f '(t)=2(4t + 8)^-1/2
f ''(t)=-4(4t + 8)^-3/2
f '''(t)=24(4t + 8)^-5/2
f'''(1/4)=24(4(1/4) + 8)^-5/2
Ans= 24/243

f '(t) = 2/√(4t + 8)

f "(t) = -4/(4t + 8)^(3/2)

f '''(t) = 24/(4t + 8)^(5/2)

f "'(1/4) = 24/(3^5) = 8/81

24/(8 + 4 t)^(5/2)

8/81