I am trying to calculate the derivative of $f$ using the product rule and quotient rule respectively.
However, I am getting different results for the product rule and quotient rule.
Did I made any mistake along the way?
$$f(x) = \frac{(x^3 + \cos x)}{6}$$
using the product rule (multiple by 1/6 instead of divide by 6)
$$f'(x) = \frac {1}{6} \frac{d}{dx}[x^3 + \cos x]$$ $$ = \frac {1}{6}(3x^2-\sin x)$$
using the quotient rule
$$f'(x) = \frac {6 \frac {d}{dx}[x^3+\cos x]-(x^3+\cos x)\frac {d}{dx}[6]}{6^2} $$ $$ = \frac {6(3x^2-\sin x)-(x^3+\cos x)}{36}$$ $$ = \frac{(3x^2-\sin x-x^3-\cos x)}{6}$$
Note that by quotient rule we have
$$f'(x) = \frac {6 \frac {d}{dx}[x^3+\cos x]-(x^3+\cos x)\frac {d}{dx}[6]}{6^2}=\frac {6 (3x^2-\sin x)-0}{6^2}=\frac {1}{6}(3x^2-\sin x)$$