Question

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

Answer 1

`g'(t)=-(250t^3+75t^2+5)/(25t^6+10t^4+t^2)`

`g''(t)=(10(375t^5+150t^4-25t^3+15t^2+1))/(t^3(5t^2+1)^3)`

Explanation:

Immediately, I want to eliminate all fractions. To do this, I'll multiply `g(t)` by `(5t)/(5t)`.

`g(t)=(5+1/t)/(t^2+1/5)((5t)/(5t))=(25t+5)/(5t^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(t)=(f(t))/(h(t))`

Then its derivative equals

`g'(t)=(h(t)f'(t)-f(t)h'(t))/(h(t))^2`

So, here, we have

`f(t)=25t+5" "=>" "f'(t)=25`

`h(t)=5t^3+t" "=>" "h'(t)=15t^2+1`

This gives us a derivative of

`g'(t)=((5t^3+t)(25)-(25t+5)(15t^2+1))/(5t^3+t)^2`

Now, simply expand.

`g'(t)=(125t^3+25t-375t^3-75t^2-25t-5)/(5t^3+t)^2`

`g'(t)=(-250t^3-75t^2-5)/(5t^3+t)^2`

This can also be written as

`g'(t)=-(250t^3+75t^2+5)/(25t^6+10t^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''(t)=-((25t^6+10t^4+t^2)(750t^2+150t)-(250t^3+75t^2+5)(150t^5+40t^3+2t))/(25t^6+10t^4+t^2)^2`

The correct final answer should be

`g''(t)=(10(375t^5+150t^4-25t^3+15t^2+1))/(t^3(5t^2+1)^3)`