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anonymous
 3 years ago
Easiest way to go about taking this derivative?
Power rule, chain rule and quotient rule is kind of ugly.
anonymous
 3 years ago
Easiest way to go about taking this derivative? Power rule, chain rule and quotient rule is kind of ugly.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[f(x) =\sqrt{\frac{ x1}{ x+1}}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0chain rule first then quotient rule

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and it isnt ugly at all. The higher u go, the more 'uglier' it gets. dont be afraid of 'ugly' solutions

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oops supposed to be 1x and 1+x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0still doesnt matter just flip the sings

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353220538034:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0chain rule the exponent 1/2 then differentiate the one inside the bracket using quotient rule

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353220581666:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it really isnt ugly at all... i have seen even uglier ones at my level

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just make the exponent 1/2 into a square root and put it beside the (1+x) ^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0same with the 2 in the denominator

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is for a first derivative analysis... It is ugly to find zeroes.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353220820451:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353220855332:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what are you trying to find?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0first and second derivative analysis

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for turning points and maximum/minimum?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0max, min, inflections

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there is a min at x=1
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