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2. V(t) = t + 1 / t + 4

(https://biology-forums.com/gallery/42/6_15_02_21_1_43_36.png) (https://biology-forums.com/index.php?action=gallery;sa=view&id=42522)

can u explain first step? I thought you r supposed to find lcd, t(t + h)...

Hi Orlando0 (https://biology-forums.com/index.php?action=profile;u=982449)

It's like adding fractions. I used the same technique used in this video, specifically when I combine the numerator fractions together, have a watch:

https://www.youtube.com/watch?v=qmTDfgpjF5w

just want to make sure but t + 1 times t = 2t or is it t(t + 1)? I'm just checking something

Will answer when I get home!

Could you highlight the part of my solution that this represents?

I still dont really get this question. I got other fraction type questions right using what I always did. this is what i did. what would the lcd be by multiplying these??

Hi Orlando0 (https://biology-forums.com/index.php?action=profile;u=982449)

Sorry, but I don't understand what you're asking. Do you mind putting an arrow at the step where you got confused in my solution?

https://biology-forums.com/index.php?action=gallery;sa=view&id=42522

I dont understand these two steps. please explain

The first step is when I applied \((t + h)\) into the function \(v(t)\).

Remember the definition of a derivative is:

\(\lim _{h\rightarrow 0}\frac{v\left(t+h\right)-v\left(t\right)}{h}\)

\(\lim _{h\rightarrow 0}\frac{\frac{\left(t+h\right)+1}{\left(t+h\right)+4}-\frac{t+1}{t+4}}{h}\)

So far so good?

oh ok yes

Shall I continue?

yes

Focus now on the numerator part:

\(\frac{\left(t+h\right)+1}{\left(t+h\right)+4}-\frac{t+1}{t+4}\)

To combine these two terms, multiply the denominators without expanding, and cross-multiply each denominator with its opposite numerator:

\(\frac{\left[\left(t+h\right)+1\right]\left(t+4\right)-\left(t+1\right)\left[\left(t+h\right)+4\right]}{\left[\left(t+h\right)+4\right]\left(t+4\right)}\)

Clean it up a little:

\(\frac{\left(t+h+1\right)\left(t+4\right)-\left(t+1\right)\left(t+h+4\right)}{\left(t+h+4\right)\left(t+4\right)}\)

Now place it back into the limit:

\(\lim _{h\rightarrow 0}\frac{\frac{\left(t+h+1\right)\left(t+4\right)-\left(t+1\right)\left(t+h+4\right)}{\left(t+h+4\right)\left(t+4\right)}}{h}\)

Simplify further:

\(\lim _{h\rightarrow 0}\frac{\left(t+h+1\right)\left(t+4\right)-\left(t+1\right)\left(t+h+4\right)}{\left(t+h+4\right)\left(t+4\right)h}\)

Now you know how I got those first two steps!

thanks so much better now