Question

I disagree with my textbook. Need help

Answer 1

whatever it is, i am on your side

Answer 2

So on the pic, under the POWER RULE step

Answer 3

can't see it though, screen shot is bad

Answer 4

Where did they take in account for the - 1/2 exponent that was supposed to be there when they took derivative?

Answer 5

Its working for me @satellite73

Answer 6

Can anyone see it?

Answer 7

what book is it?

Answer 8

Thomas' calculus Alternete edition

Answer 9

can anyone see the pic?

Answer 10

cant find the book sorry!! :( Try taking a better pic.

Answer 11

Gurr. Give me time and I will type it out

Answer 12

\[\ln (x^2+1)+\ln(x+3)^{1/2}-\ln(x-1)\]
Then
\[\ln (x^2+1)+1/2 \ln (x+3)- \ln (x-1)\]

Answer 13

The idea is \[
\ln\left((x+3)^{1/2}\right) = \frac 12 \ln(x+3)
\]

Answer 14

How?

Answer 15

http://www.wolframalpha.com/input/?i=ln+*+%28x%2B3%29%5E%281%2F2%29

Answer 16

\[\begin{array}{rcl}
\ln\left((x+3)^{1/2}\right) &=& y\\
(x+3)^{1/2} &=& e^y\\
x+3 &=& (e^y)^2\\
x+3 &=& e^{2y} \\
\ln(x+3) &=& 2y \\
\frac 12 \ln(x+3) &=& y \\
\ln\left((x+3)^{1/2}\right) &=& \frac 12 \ln(x+3)
\end{array}
\]

Answer 17

Ok. Thanks. Im satisfied now