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Differentiate f(x)= (x-6)^2/3 where c=6 using the alternative of the derivative
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alternative form: f(x)-f(c)/ x-c as limit of x approaches c
Is the problem\[(x-6)^{2/3}\] or \[\frac{(x-6)^{2} }{3}\]
oh sorry it's the first one (x-6)^(2/3)
Okay brb working on problem.
lol kk
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\[f'(x)= \frac{f(x)-f(c)}{x-c}\] Where c=6 f(6)=0 so then we have \[f'(x)= \frac{(x-6)^{2/3}-0}{x-6}\] or \[f'(x)= \frac{(x-6)^{2/3}}{x-6}\]
does that come down to 1/ x-6?
Using a different method, the correct answer is (2/3)(x-6)^(-1/3) \[\frac{2}{3}(x-6)^{-1/3}\] I have to go, but I can help you in like ~40 mins if you still have problems
You can't simplify to 1/(x-6) because of the (3/2) power in the numerator.
oh alrite thank you
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