AmTran_Bus
  • AmTran_Bus
I disagree with my textbook. Need help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
whatever it is, i am on your side
AmTran_Bus
  • AmTran_Bus
AmTran_Bus
  • AmTran_Bus
So on the pic, under the POWER RULE step

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anonymous
  • anonymous
can't see it though, screen shot is bad
AmTran_Bus
  • AmTran_Bus
Where did they take in account for the - 1/2 exponent that was supposed to be there when they took derivative?
AmTran_Bus
  • AmTran_Bus
Its working for me @satellite73
AmTran_Bus
  • AmTran_Bus
Can anyone see it?
anonymous
  • anonymous
what book is it?
AmTran_Bus
  • AmTran_Bus
Thomas' calculus Alternete edition
AmTran_Bus
  • AmTran_Bus
can anyone see the pic?
pooja195
  • pooja195
cant find the book sorry!! :( Try taking a better pic.
AmTran_Bus
  • AmTran_Bus
Gurr. Give me time and I will type it out
AmTran_Bus
  • AmTran_Bus
\[\ln (x^2+1)+\ln(x+3)^{1/2}-\ln(x-1)\] Then \[\ln (x^2+1)+1/2 \ln (x+3)- \ln (x-1)\]
anonymous
  • anonymous
The idea is \[ \ln\left((x+3)^{1/2}\right) = \frac 12 \ln(x+3) \]
AmTran_Bus
  • AmTran_Bus
How?
AmTran_Bus
  • AmTran_Bus
http://www.wolframalpha.com/input/?i=ln+*+%28x%2B3%29%5E%281%2F2%29
anonymous
  • anonymous
\[\begin{array}{rcl} \ln\left((x+3)^{1/2}\right) &=& y\\ (x+3)^{1/2} &=& e^y\\ x+3 &=& (e^y)^2\\ x+3 &=& e^{2y} \\ \ln(x+3) &=& 2y \\ \frac 12 \ln(x+3) &=& y \\ \ln\left((x+3)^{1/2}\right) &=& \frac 12 \ln(x+3) \end{array} \]
AmTran_Bus
  • AmTran_Bus
Ok. Thanks. Im satisfied now

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