# Thread: Constant Multiple Rule...Yes or No?

1. ## Constant Multiple Rule...Yes or No?

Find s'(t).

Does the constant multiple rule apply here?

s(t) = [-2(2-t) * sqrt{1+t}]/3

I pulled out 1/3 but my answer is nothing like the book's answer.

Applying the constant rule, the problem now looks like this:

s(t) = (1/3) times [-2(2-t) * sqrt{1+t}]

The fraction (1/3) comes along for the ride.
I then know to apply the product rule to everything else.

The answer is (t)/sqrt{1+t}. Can someone work out this problem for me? I would love to see the steps. I never post a question in this forum without working out the problem many times on my own. Please, help. Thank you.

2. ## Re: Constant Multiple Rule...Yes or No?

Yes the constant multiple rule applies here, in fact, it would be easiest if you take out -2/3 as a factor.

3. ## Re: Constant Multiple Rule...Yes or No?

I will try this on my own one more time. I never like giving up.

4. ## Re: Constant Multiple Rule...Yes or No?

(-t)/[2(sqrt{1+t})

(t)/sqrt{1+t}

I came close but not right. Can you solve this one step by step? Thanks again.

5. ## Re: Constant Multiple Rule...Yes or No?

Ok, so you apply the product rule to $\displaystyle -2(2-t)\sqrt{1+t}$.

You should get $\displaystyle \sqrt{1+t}\frac{d}{dt}(-2(2-t))-2(2-t)\frac{d}{dt}\sqrt{1+t}$. The first term you can multiply top and bottom by $\displaystyle \sqrt{1+t}$ to get $\displaystyle \sqrt{1+t}$ in the denominator, and the second will have it when you take the derivative. So what does each term have in the numerator?

- Hollywood

6. ## Re: Constant Multiple Rule...Yes or No?

I'll try this one again tomorrow (my day off). I am going chapter by chapter in my single variable calculus text which covers calculus 1 and 2. I always wanted to learn calculus 1-3. Why take out a loan and run myself into debt?

The internet makes it possible for me to learn calculus.
If I can go as far as mutivariable calculus on my own and learn it sufficiently well, it is a great accomplishment. The next chapter is Implicit Differentiation.