Finding Derivatives

**VitaX** · September 23rd, 2009, 05:31 PM

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.

**Calculus26** · September 23rd, 2009, 05:47 PM

dv/dt = 1 + 1/t^2

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

how did you get -2/t ?

**VitaX** · September 23rd, 2009, 06:31 PM

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?

**Calculus26** · September 23rd, 2009, 06:50 PM

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

**Calculus26** · September 23rd, 2009, 07:01 PM

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

**VitaX** · September 23rd, 2009, 07:04 PM

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?

**Calculus26** · September 23rd, 2009, 07:06 PM

right click on the attachment and click on open link

**Calculus26** · September 23rd, 2009, 07:08 PM

of course I meant click on "open link"

**VitaX** · September 23rd, 2009, 07:11 PM

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.

**Calculus26** · September 23rd, 2009, 07:14 PM

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

mistakes

**VitaX** · September 23rd, 2009, 07:44 PM

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

**Calculus26** · September 23rd, 2009, 07:48 PM

Let h = 0

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

**pham07** · September 23rd, 2009, 07:51 PM

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

**Calculus26** · September 23rd, 2009, 07:56 PM

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?

**VitaX** · September 23rd, 2009, 07:59 PM

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

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