1. Finding Derivatives

Find the derivative:

$\frac{dv}{dt}$ if $v=t-\frac{1}{t}$

My answer is $-\frac{2}{t}$

I can show work later if needed if this is wrong. Just need to be sure I did this one right before I go on and do others.

2. dv/dt = 1 + 1/t^2

since dt/dt = 1 and d(1/t)/dt = -1/t^2

how did you get -2/t ?

3. By definition $\frac{f(x+h) - f(x)}{h}$ as h goes to 0.I first changed $t - \frac{1}{t} = \frac{t^2 - 1}{t}$. Plugged that in and got $\frac{1}{h}[(\frac{(t+h)^2 - 1}{t+h} - (\frac{t^2 - 1}{t})]$. Common denominator is $t(t + h)$. Then it becomes $\frac{1}{h}[\frac{t^2 + 2ht + h^2 - 1 - t^2 - 1}{t(t + h)}]$. Then $\frac{1}{h}[\frac{h^2 + 2ht - 2}{t(t + h)}]$. Then $\frac{h^2 + 2ht - 2}{ht(t + h)}$. Then $\frac{h + 2 - 2}{t + h}$ Substitute $0$ in for $h$ and you are left with $-\frac{2}{t}$

Edit: Looks like 2 crosses out in the numerator so you are left with $0$ ... hows this?

4. See attachment it is easier if you don't combine

By the way in my original post I ha a neg sign wrong but have corrected it

5. By the way if you do combine you have what is in the following attachment

6. When I click the attachments my screen just goes dark. Nothing appears. Maybe an error on my pc in loading it. But I found a mistake in my work and am left with $\frac{0}{t}$ which is just $0$. Hows that?

7. right click on the attachment and click on open link

8. of course I meant click on "open link"

9. So where did I go wrong in my post above?

Edit I think I made a sign error moving all under one denominator. So I think the answer is $\frac{2}{t}$ if not then I'm completely lost as I did this problem the same basic way as my teacher showed us in class.

10. See my second attachment--It is more helpful for you if you find your own

mistakes

11. Scratched this post. I understand everything you did and my mistakes. Thanks for your help.

12. Let h = 0

you get (t^2 + 1)/(t^2) = 1 + 1/t^2

13. you are trying to cancel terms that you can't cancel. since the top is a sum and not a product, you can't cancel out the "t's".. You will get (t^2+1)/t^2, which is the same as writing

(t^2/t^2)+ 1/t^2

14. you are trying to cancel terms that you can't cancel. since the top is a sum and not a product, you can't cancel out the "t's".. You will get (t^2+1)/t^2, which is the same as writing

(t^2/t^2)+ 1/t^2

which simplifies to?

15. Yah I see it thats why I deleted my post above as it was dumb of me not to see it.