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

Jun 2, 2015

We must use the quotient rule, which states that

Be $y = f \frac{x}{g} \left(x\right)$, then $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

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

• $f \left(t\right) = {t}^{2} + 2$
• $f ' \left(t\right) = 2 t$

• $g \left(t\right) = {\left(6 t - 3\right)}^{7}$

• $g ' \left(t\right)$ demands chain rule, which states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$, so $g ' \left(t\right) = \left(7 {u}^{6}\right) \cdot 6 = \textcolor{g r e e n}{42 {\left(6 t - 3\right)}^{6}}$

Thus,

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2 t {\left(6 t - 3\right)}^{7} - \left({t}^{2} + 2\right) \left(42 {\left(6 t - 3\right)}^{6}\right)}{6 t - 3} ^ 14 =$

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

Considering ${\left(6 t - 3\right)}^{8}$ our l.c.d.:

$\frac{2 t \left(6 t - 3\right) - 42 \left({t}^{2} + 2\right)}{6 t - 3} ^ 8 = \frac{12 {t}^{2} - 6 t - 42 {t}^{2} - 84}{6 t - 3} ^ 8 = \textcolor{G r e e n}{\frac{- 30 {t}^{2} - 6 t - 84}{6 t - 3} ^ 8}$