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

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baldymcgee6Best ResponseYou've already chosen the best response.0
\[f(x) =\sqrt{\frac{ x1}{ x+1}}\]
 one year ago

irkizBest ResponseYou've already chosen the best response.0
chain rule first then quotient rule
 one year ago

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

baldymcgee6Best ResponseYou've already chosen the best response.0
oops supposed to be 1x and 1+x
 one year ago

irkizBest ResponseYou've already chosen the best response.0
still doesnt matter just flip the sings
 one year ago

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

baldymcgee6Best ResponseYou've already chosen the best response.0
dw:1353220581666:dw
 one year ago

irkizBest ResponseYou've already chosen the best response.0
it really isnt ugly at all... i have seen even uglier ones at my level
 one year ago

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

irkizBest ResponseYou've already chosen the best response.0
same with the 2 in the denominator
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
This is for a first derivative analysis... It is ugly to find zeroes.
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
dw:1353220855332:dw
 one year ago

irkizBest ResponseYou've already chosen the best response.0
what are you trying to find?
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
first and second derivative analysis
 one year ago

irkizBest ResponseYou've already chosen the best response.0
for turning points and maximum/minimum?
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
max, min, inflections
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
there is a min at x=1
 one year ago
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