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

##### 1 Answer

#### Explanation:

Immediately, I want to eliminate all fractions. To do this, I'll multiply

#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:

The correct final answer should be

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