Grove HSC 3U chapter 4.2 Question 2 - Differentiating an exponential type fn.

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Question 2. =============

$y=(e^{2x}+1)^7$
$y'=7.(e^{2x}+1)^6\times 2e^{2x}$ by chain rule.
$y'=14(e^{2x}+1)^6e^{2x}$
$y'=14 e^{2x} (e^{2x}+1)^6$

Now use product rule too for $y''$

$u=14e^{2x}$, $v=(e^{2x}+1)^6$
$u'=28e^{2x}$, $v'=6(e^{2x}+1)^5 2e^{2x}=12 e^{2x}(e^{2x}+1)^5$

$y'=14e^{2x}12 e^{2x}(e^{2x}+1)^5 + 28e^{2x}(e^{2x}+1)^6$

Factorising

$y'=28e^{2x}(e^{2x}+1)^5(7e^{2x}+1)$