# How do you find the first and second derivatives of g(t)=(5+1/t)/(t^2+1/5)  using the quotient rule?

Mar 26, 2016

$g ' \left(t\right) = - \frac{250 {t}^{3} + 75 {t}^{2} + 5}{25 {t}^{6} + 10 {t}^{4} + {t}^{2}}$

$g ' ' \left(t\right) = \frac{10 \left(375 {t}^{5} + 150 {t}^{4} - 25 {t}^{3} + 15 {t}^{2} + 1\right)}{{t}^{3} {\left(5 {t}^{2} + 1\right)}^{3}}$

#### Explanation:

Immediately, I want to eliminate all fractions. To do this, I'll multiply $g \left(t\right)$ by $\frac{5 t}{5 t}$.

$g \left(t\right) = \frac{5 + \frac{1}{t}}{{t}^{2} + \frac{1}{5}} \left(\frac{5 t}{5 t}\right) = \frac{25 t + 5}{5 {t}^{3} + t}$

Now, we can apply the quotient rule, which states that if we have a function that is a quotient of two functions, such as

$g \left(t\right) = \frac{f \left(t\right)}{h \left(t\right)}$

Then its derivative equals

$g ' \left(t\right) = \frac{h \left(t\right) f ' \left(t\right) - f \left(t\right) h ' \left(t\right)}{h \left(t\right)} ^ 2$

So, here, we have

$f \left(t\right) = 25 t + 5 \text{ "=>" } f ' \left(t\right) = 25$

$h \left(t\right) = 5 {t}^{3} + t \text{ "=>" } h ' \left(t\right) = 15 {t}^{2} + 1$

This gives us a derivative of

$g ' \left(t\right) = \frac{\left(5 {t}^{3} + t\right) \left(25\right) - \left(25 t + 5\right) \left(15 {t}^{2} + 1\right)}{5 {t}^{3} + t} ^ 2$

Now, simply expand.

$g ' \left(t\right) = \frac{125 {t}^{3} + 25 t - 375 {t}^{3} - 75 {t}^{2} - 25 t - 5}{5 {t}^{3} + t} ^ 2$

$g ' \left(t\right) = \frac{- 250 {t}^{3} - 75 {t}^{2} - 5}{5 {t}^{3} + t} ^ 2$

This can also be written as

$g ' \left(t\right) = - \frac{250 {t}^{3} + 75 {t}^{2} + 5}{25 {t}^{6} + 10 {t}^{4} + {t}^{2}}$

I'll leave the work of finding the second derivative to the reader, since it becomes more menial than practical, but here's how it should be approached:

$g ' ' \left(t\right) = - \frac{\left(25 {t}^{6} + 10 {t}^{4} + {t}^{2}\right) \left(750 {t}^{2} + 150 t\right) - \left(250 {t}^{3} + 75 {t}^{2} + 5\right) \left(150 {t}^{5} + 40 {t}^{3} + 2 t\right)}{25 {t}^{6} + 10 {t}^{4} + {t}^{2}} ^ 2$

The correct final answer should be

$g ' ' \left(t\right) = \frac{10 \left(375 {t}^{5} + 150 {t}^{4} - 25 {t}^{3} + 15 {t}^{2} + 1\right)}{{t}^{3} {\left(5 {t}^{2} + 1\right)}^{3}}$