<s>[INST] Derivative of $f(x) = \frac{x^3+\cos x}{6}$ 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
$$f'(x) = \frac {6 \frac {d}{dx}[x^3+\cos x]-(x^3+\cos x)\frac {d}{dx}[6]}{6^2} $$
$$      = \frac {6(3x^2-\sin x)-(x^3+\cos x)}{36}$$
$$      = \frac{(3x^2-\sin x-x^3-\cos x)}{6}$$        
 [/INST] Note that by quotient rule we have
$$f'(x) = \frac {6 \frac {d}{dx}[x^3+\cos x]-(x^3+\cos x)\frac {d}{dx}[6]}{6^2}=\frac {6 (3x^2-\sin x)-0}{6^2}=\frac {1}{6}(3x^2-\sin x)$$
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