solving using definition of derivitave...

[sydney_bristow87]

Alrighty, so I get my exam back, we go over the problems we got wrong, and I write down the steps to the correct solution so I can study for the retake (we're retaking our midterm, does that say something to you?) ...so, today, I'm going through my notes, and I understand everything, except this stupid thing with fractions... I have NO idea what happened between two of the steps...

Solve using the definition of a derivative:
\(\displaystyle f(t)= \frac{2}{t}\)

So, we have the definition of a derivative:

\(\displaystyle \[ \lim_{t\to 0} \frac{f(t + \Delta t) - f(t)}{\Delta t}]\)

and we plug everything in:

\(\displaystyle \frac{\frac{2}{t + \Delta t} - \frac{2}{t}}{\Delta t}\)

then, we make a common denominator, by multiplying \(\displaystyle \frac{2}{t}\) by \(\displaystyle t + \Delta t\) and by multiplying \(\displaystyle \frac{2}{t + \Delta t}\) by \(\displaystyle \frac{2}{t}\)

this gives us:

\(\displaystyle \frac{\frac{2t - 2(t + \Delta t)}{t(t + \Delta t)}}{\Delta t}\)

This is where I get completely LOST. In my notes, the above equation is muliplied by \(\displaystyle \frac{\Delta t}{1}\), but I'm not entirely sure if I copied it down right. Either way, it's between this step and the next that I have no idea what I did... In my notes, after \(\displaystyle \frac{\frac{2t - 2(t + \Delta t)}{t(t + \Delta t)}}{\Delta t}\) is multiplied by \(\displaystyle \frac{\Delta t}{1}\), it turns into this:

\(\displaystyle \frac{-2 \Delta t}{(\Delta t)(t)(t + \Delta t)}\)

the \(\displaystyle \Delta t\)'s cancel each other out, so we end up with:

\(\displaystyle \frac{-2}{t (t + \Delta t)}\)

The limit of 0 is applied, and the solution is:

\(\displaystyle \frac{-2}{t^2}\)

Now, my problem is, I have NO IDEA how I got from here:

\(\displaystyle \frac{\frac{2t - 2(t + \Delta t)}{t(t + \Delta t)}}{\Delta t}\) times \(\displaystyle \frac{\Delta t}{1}\)

to

\(\displaystyle \frac{-2 \Delta t}{(\Delta t)(t)(t + \Delta t)}\)

does anyone know?

 

[galactus]

It looks like you're just having algebra troubles.

I am going to use h instead of delta t.

We have:

\(\displaystyle \frac{\frac{2}{t+h}-\frac{2}{t}}{h}\)

=\(\displaystyle \frac{2}{th+h^{2}}-\frac{2}{th}\)

=\(\displaystyle \frac{2th-2th-2h^{2}}{th(th+h^{2})}\)

=\(\displaystyle \frac{-2h^{2}}{t^{2}h^{2}+th^{3}}\)

=\(\displaystyle \frac{-2h^{2}}{t^{2}h^{2}}+\frac{2h^{2}}{th^{3}}\)

=\(\displaystyle \frac{-2}{t^{2}}+\frac{2}{th}\)

As h approaches 0 you can see the limit approaches \(\displaystyle \frac{-2}{t^{2}}\)

Almost forgot. As for what you done:

\(\displaystyle \frac{\frac{2t-2(t+{\Delta}t)}{t(t+{\Delta}t)}}{{\Delta}t}\)

=\(\displaystyle \frac{2t-2(t+{\Delta}t)}{t(t+{\Delta}t)}\frac{1}{{\Delta}t}\)

=\(\displaystyle \frac{2t-2t-2{\sout{\Delta}}t}{t(t+{\Delta}t)}\frac{1}{\sout{{\Delta}t}}\)

=\(\displaystyle \frac{-2}{t(t+{\Delta}t)}\)

As \(\displaystyle {\Delta}t\) approaches 0:

\(\displaystyle \frac{-2}{t^{2}+0}=\frac{-2}{t^{2}}\)

 

[Daniel_Feldman]

Alrighty, so I get my exam back, we go over the problems we got wrong, and I write down the steps to the correct solution so I can study for the retake (we're retaking our midterm, does that say something to you?) ...so, today, I'm going through my notes, and I understand everything, except this stupid thing with fractions... I have NO idea what happened between two of the steps...

Solve using the definition of a derivative:
\(\displaystyle f(t)= \frac{2}{t}\)

So, we have the definition of a derivative:

\(\displaystyle \[ \lim_{t\to 0} \frac{f(t + \Delta t) - f(t)}{\Delta t}]\)

and we plug everything in:

\(\displaystyle \frac{\frac{2}{t + \Delta t} - \frac{2}{t}}{\Delta t}\)

then, we make a common denominator, by multiplying \(\displaystyle \frac{2}{t}\) by \(\displaystyle t + \Delta t\) and by multiplying \(\displaystyle \frac{2}{t + \Delta t}\) by \(\displaystyle \frac{2}{t}\)

this gives us:

\(\displaystyle \frac{\frac{2t - 2(t + \Delta t)}{t(t + \Delta t)}}{\Delta t}\)

This is where I get completely LOST. In my notes, the above equation is muliplied by \(\displaystyle \frac{\Delta t}{1}\), but I'm not entirely sure if I copied it down right. Either way, it's between this step and the next that I have no idea what I did... In my notes, after \(\displaystyle \frac{\frac{2t - 2(t + \Delta t)}{t(t + \Delta t)}}{\Delta t}\) is multiplied by \(\displaystyle \frac{\Delta t}{1}\), it turns into this:

\(\displaystyle \frac{-2 \Delta t}{(\Delta t)(t)(t + \Delta t)}\)

the \(\displaystyle \Delta t\)'s cancel each other out, so we end up with:

\(\displaystyle \frac{-2}{t (t + \Delta t)}\)

The limit of 0 is applied, and the solution is:

\(\displaystyle \frac{-2}{t^2}\)

Now, my problem is, I have NO IDEA how I got from here:

\(\displaystyle \frac{\frac{2t - 2(t + \Delta t)}{t(t + \Delta t)}}{\Delta t}\) times \(\displaystyle \frac{\Delta t}{1}\)

to

\(\displaystyle \frac{-2 \Delta t}{(\Delta t)(t)(t + \Delta t)}\)

does anyone know?

Here's what you did:

To get from:

\(\displaystyle \frac{\frac{2t - 2(t + \Delta t)}{t(t + \Delta t)}}{\Delta t}\)

to

\(\displaystyle \frac{-2 \Delta t}{(\Delta t)(t)(t + \Delta t)}\)


you actually multiplied by

1/delta(t) rather than delta(t)/1

 

[sydney_bristow87]

=/

I got at least three problems wrong on the exam because I messed up with fractions... I think I need remedial third grade math...

you actually multiplied by

1/delta(t) rather than delta(t)/1

But wouldn't that make the denominator \(\displaystyle \Delta t^2\)?

Almost forgot. As for what you done:

\(\displaystyle \frac{\frac{2t-2(t+{\Delta}t)}{t(t+{\Delta}t)}}{{\Delta}t}\)

=\(\displaystyle \frac{2t-2(t+{\Delta}t)}{t(t+{\Delta}t)}\frac{1}{{\Delta}t}\)

=\(\displaystyle \frac{2t-2t-2{\sout{\Delta}}t}{t(t+{\Delta}t)}\frac{1}{\sout{{\Delta}t}}\)

=\(\displaystyle \frac{-2}{t(t+{\Delta}t)}\)

As \(\displaystyle {\Delta}t\) approaches 0:

\(\displaystyle \frac{-2}{t^{2}+0}=\frac{-2}{t^{2}}\)

I see what you did there, but again, how would multiplying by 1/delta t get rid of the delta t on the denominator? Aside from that, the problem makes more sense to me now - I see what's going on with the numerator...

Still can't see how my teacher got to: \(\displaystyle \frac{-2 \Delta t}{\Delta t (t)(t + \Delta t)}\)

I hate fractions... my teacher had to explain one problem about three times because I couldn't figure out where the numerator had gone - actually, that might have been this one he explained... =/

 

[sydney_bristow87]

Nevermind It clicked - you mulitply by 1/ delta t on both the numerator AND the denominator - that way, you get delta t / delta t, which = 1, therefore getting rid of the denominator...

Yeah, I really think I do need some kind of fraction review.