$\displaystyle f(x) = \frac{2}{3} \sqrt{36-p^2)}$
$\displaystyle f'(x) =\frac{1}{2}(36-p^2)^{-\frac{1}{2}} \cdot -2p$
Am i doing this right? Im a bit lost.
Lol.
Anyway, he or she is asking a valid point (although it really doesn't have much an impact at this stage of game): your original function is in terms of x, but the variable you are differentiating is "p". Is it actually F(p)?
As for your process, square root is the power of 1/2 - what is being asked, is where did the 2/3 go? You don't need to factor is, just multiply it by 1/2 (and then simplify some more with the -2p).
Ok 2 Things:
First, VonNemo19 said, if you are taking the derivative with respect to x for a function that does not have an x in it, then the whole thing is zero (since it can be treated like a constant). But then again, it depends on your instructions.
Second, assuming it's actually f'(p) and no f'(x), then you are correct to bring down the 1/2, but you are incorrect to remove the 2/3 since it is in the same term (via multiplication). So the 1/2 and 2/3 should cancel for the twos, and you should end up with 1/3.
The correct answer would be:
$\displaystyle
-\frac{2}{3}x(36-x^2)^{-\frac{1}{2}}
$
Agree to disagree. He or she has more than enough time to learn proper Math-English (assuming they did make a typo - which I'm pretty sure they did, as I never saw any problem in Calc 1 that would have been sneaky like that). But I'm not interested in getting into a pissing match here.
Calc 1 has implicit differentiation, correct? Or was that in Calculus 2? Regardless, if this were an implicit differentiation problem, the answer could be signficantly different if there is more information in the instructions.
On the other hand, an error was made in the differentiation no matter what the instructions were (no instructions would cause the 2/3 to disappear like that), and so obviously that is something which should be addressed.