Grove HSC 2U Ex 4.2 Q2o

Question= 

Answer =

How did they get that?

Thanks, Anna

SOLUTION

$u=e^{2x+1}$, $u'=2e^{2x+1}$
$v=2x+5$, $v'=2$

$y'= \displaystyle { vu'-uv' \over v^2}$
$y'= \displaystyle { (2x+5)2e^{2x+1}-e^{2x+1}\times 2 \over (2x+5)^2}$
$y'= \displaystyle {  2e^{2x+1}(2x+5-1)\over (2x+5)^2}$
$y'= \displaystyle {  2e^{2x+1}(2x+4)\over (2x+5)^2}$

$y'= \displaystyle {  2\times 2e^{2x+1}(x+2)\over (2x+5)^2}$
$y'= \displaystyle {  4e^{2x+1}(x+2)\over (2x+5)^2}$

:-)

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