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

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

  • 6
    $\begingroup$ The derivative of a constant is $0$, not $1$. $\endgroup$ – Arnaud Mortier May 22 '18 at 14:19
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    $\begingroup$ Was this an excercise ? Seems unecessary to use the quotient rule. $\endgroup$ – XPenguen May 22 '18 at 14:24
  • $\begingroup$ @ArnaudMortier Thank you for pointing out my careless mistake! $\endgroup$ – ilovetolearn May 22 '18 at 14:26
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    $\begingroup$ @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.) $\endgroup$ – Barry Cipra May 22 '18 at 14:33
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    $\begingroup$ @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. $\endgroup$ – Barry Cipra May 22 '18 at 14:42

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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