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AmTran_Bus
 one year ago
I disagree with my textbook. Need help
AmTran_Bus
 one year ago
I disagree with my textbook. Need help

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0whatever it is, i am on your side

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0So on the pic, under the POWER RULE step

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can't see it though, screen shot is bad

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0Where did they take in account for the  1/2 exponent that was supposed to be there when they took derivative?

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0Its working for me @satellite73

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0Thomas' calculus Alternete edition

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0can anyone see the pic?

pooja195
 one year ago
Best ResponseYou've already chosen the best response.0cant find the book sorry!! :( Try taking a better pic.

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0Gurr. Give me time and I will type it out

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0\[\ln (x^2+1)+\ln(x+3)^{1/2}\ln(x1)\] Then \[\ln (x^2+1)+1/2 \ln (x+3) \ln (x1)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The idea is \[ \ln\left((x+3)^{1/2}\right) = \frac 12 \ln(x+3) \]

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=ln+*+%28x%2B3%29%5E%281%2F2%29

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{array}{rcl} \ln\left((x+3)^{1/2}\right) &=& y\\ (x+3)^{1/2} &=& e^y\\ x+3 &=& (e^y)^2\\ x+3 &=& e^{2y} \\ \ln(x+3) &=& 2y \\ \frac 12 \ln(x+3) &=& y \\ \ln\left((x+3)^{1/2}\right) &=& \frac 12 \ln(x+3) \end{array} \]

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0Ok. Thanks. Im satisfied now
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