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

$$f'(x) = \frac {6 \frac {d}{dx}[x^3+\cos x]-(x^3+\cos x)\frac {d}{dx}[6]}{6^2}$$ $$= \frac {6(3x^2-\sin x)-(x^3+\cos x)}{36}$$ $$= \frac{(3x^2-\sin x-x^3-\cos x)}{6}$$

• The derivative of a constant is $0$, not $1$. – Arnaud Mortier May 22 '18 at 14:19
• Was this an excercise ? Seems unecessary to use the quotient rule. – XPenguen May 22 '18 at 14:24
• @ArnaudMortier Thank you for pointing out my careless mistake! – ilovetolearn May 22 '18 at 14:26
• @youcanlearnanything, in addition to the key mistake pointed out by Arnaud Mortier, you also cancelled a $6$ incorrectly in the final step. (The $6$ in the numerator only applies to the first set of parentheses.) – Barry Cipra May 22 '18 at 14:33
• @youcanlearnanything, I noticed the error in part because I (almost) made a similar mistake earlier today (meaning I made it but then caught it). If you are aware of a tendency toward certain kinds of mistakes, you can learn to be extra careful when doing those kinds of calculations. – Barry Cipra May 22 '18 at 14:42

$$f'(x) = \frac {6 \frac {d}{dx}[x^3+\cos x]-(x^3+\cos x)\frac {d}{dx}[6]}{6^2}=\frac {6 (3x^2-\sin x)-0}{6^2}=\frac {1}{6}(3x^2-\sin x)$$