Thread: Finding Derivatives

Find the derivative:

if

My answer is

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.

dv/dt = 1 + 1/t^2

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

how did you get -2/t ?

By definition as h goes to 0.I first changed . Plugged that in and got . Common denominator is . Then it becomes . Then . Then . Then Substitute in for and you are left with

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

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

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

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 which is just . Hows that?

right click on the attachment and click on open link

of course I meant click on "open link"

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 if not then I'm completely lost as I did this problem the same basic way as my teacher showed us in class.

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

mistakes

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

Let h = 0

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

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

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?

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