How do you differentiate $(t^2+2)/(6t-3)^7$ ?

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Answer 1 · 2015-06-02T00:42:28

We must use the quotient rule, which states that

Be $y=f(x)/g(x)$ , then $(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/(g(x))^2$

Now, before starting, we can acknowledge all our four needed functions:

$f(t)=t^2+2$
$f'(t)=2t$
$g(t)=(6t-3)^7$
$g'(t)$ demands chain rule, which states that $(dy)/(dx)=(dy)/(du)(du)/(dx)$ , so $g'(t)=(7u^6)*6=color(green)(42(6t-3)^6)$

Thus,

$(dy)/(dt)=(2t(6t-3)^7-(t^2+2)(42(6t-3)^6))/(6t-3)^14=$

$=(2tcancel((6t-3)^7))/(6t-3)^(cancel(14)7)-((t^2+2)*42cancel((6t-3)^6))/(6t-3)^(cancel(14)8$

Considering $(6t-3)^8$ our l.c.d.:

$(2t(6t-3)-42(t^2+2))/(6t-3)^8=(12t^2-6t-42t^2-84)/(6t-3)^8=color(Green)((-30t^2-6t-84)/(6t-3)^8)$

How do you differentiate (t^2+2)/(6t-3)^7(t^2+2)/(6t-3)^7?