Question

for the function $f(x) = 6x^5 - 5x^4 + 4x^3 + 8x^2 + 1$, find $f(x)$, the third derivative of $f$, and $f^{(4)}(x)$, the fourth derivative of $f$.

$f(x) = \square$

Explanation:

Step1: Compute first derivative

$f'(x) = 6*5x^4 -5*4x^3 +4*3x^2 +8*2x = 30x^4 -20x^3 +12x^2 +16x$

Step2: Compute second derivative

$f''(x) = 30*4x^3 -20*3x^2 +12*2x +16 = 120x^3 -60x^2 +24x +16$

Step3: Compute third derivative

$f'''(x) = 120*3x^2 -60*2x +24 = 360x^2 -120x +24$

Step4: Compute fourth derivative

$f^{(4)}(x) = 360*2x -120 = 720x -120$

Answer:

$f'''(x) = 360x^2 -120x +24$
$f^{(4)}(x) = 720x -120$