Question

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

Answer 1

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)`