1. ## derivative.. problem

Hello,
I need help , can someone show me how to find the derivative of the problem in the attached photo

2. $f(t)=\left(t+\frac{1}{t}\right)^{-2}$

$f'(t) = -2 \left(t+\frac{1}{t}\right)^{-3} \cdot \left(t+\frac{1}{t}\right)'$

Can you finish from here? This is using the chain rule.

3. $\left(t+\frac{1}{t}\right)'$

Can you finish from here? This is using the chain rule.

,I'm not sure what the derivative is of this part is????
could you show me how to complete it please?
Can the quotient rule not be used..

4. Originally Posted by wolfhound
$\left(t+\frac{1}{t}\right)'$

Can you finish from here? This is using the chain rule.

,I'm not sure what the derivative is of this part is????
could you show me how to complete it please?
Can the quotient rule not be used..
$\frac{d}{dt}\left[t+\frac{1}{t}\right]=\frac{d}{dt}[t+t^{-1}]=1-t^{-2}=1-\frac{1}{t^2}$

5. Thanks for the help!
But, is this the final answer?

6. Originally Posted by wolfhound
Thanks for the help!
But, is this the final answer?
No. It is only the part that you asked for. The final answer would be:

$f'(t)=-2\left(t+\frac{1}{t}\right)^{-3}\left(1-\frac{1}{t^2}\right)$

7. Originally Posted by VonNemo19
No. It is only the part that you asked for. The final answer would be:

$f'(t)=-2\left(t+\frac{1}{t}\right)^{-3}\left(1-\frac{1}{t^2}\right)$
Thanks ,I meant was my bit in the photo the final answer,I was wrong as usual ..
so it cannot be multiplied or simplified any further?

8. Originally Posted by wolfhound
Thanks ,I meant was my bit in the photo the final answer,I was wrong as usual ..
so it cannot be multiplied or simplified any further?
Imean, sure but I know that I wouldn't mess with it anymore unless I was trying to do something with it. Maybe I would write

$f'(t)=-2\frac{\frac{t^2-1}{t^2}}{\frac{(t^2+1)^3}{t^3}}$ and simplify.

9. Oki doki man,