1. dQ/dt

I am starting with an equation:

$\displaystyle Q=\frac{at}{(1+bt^2)^3}$

to be solved for $\displaystyle \frac{dQ}{dt}$

My solutions manual has given up this hint to me:

$\displaystyle at*\frac{-6bt}{(1+bt^2)^4}+\frac{a}{(1+bt^2)^3}$

Does anyone know what rules were applied to get the equation into the form above? I looked into the quotient rule to see if I could make sense out of what happened but I don't see it yet.

2. Re: dQ/dt

Hey sepoto.

You best look at the quotient rule where d/dx (u/v) = [vu' - uv']/v^2 where u and v are functions of x.

3. Re: dQ/dt

It looks like the solution is a little bit more complex than just the application of the quotient rule. I've been working the problem but my steps don't look right to me yet.

4. Re: dQ/dt

Also I should have said earlier: also take a look at the product rule where you use d/dx (u/v) = 1/v*u' + u*(1/v)': they should give the same answer.

5. Re: dQ/dt

Chiro is right, product rule and quotient rule will both give the same answer. And the answer given is correct.

d/dx (u/v) = [vu' - uv']/v^2

u=at
u'=a

v=(1+bt^2)^3

v'= 3(1+bt^2)^2 * 2bt

Can you do it now?

Hint when you get near the end: (1-3)/4 = (1/4) - (3/4)

6. Re: dQ/dt

So by the quotient rule if I follow it then the resulting equation should be something like:

$\displaystyle \frac{(1+bt^2)^3(a)-(at)(3)(1+bt^2)^2(2bt)}{(1+bt^2)^6}$

I'm trying to factor it out and reduce it down now to what is supposed to be the answer:

$\displaystyle \frac{a(1-5bt^2)}{(1+bt^2)^4}$

I'm still working the problem to see how one might transform to the other.

7. Re: dQ/dt

Yes what you have done so far is good.
now pretty up the numerator a little and factorise out a*(1+bt^2)^2
then cancel out (1+bt^2)^2 (top and bottom)
Collect like terms and it is finished.

I don't know how they thought the hint would work I thought that was the answer that they wanted.
You have do do more work to get it in the form of the hint.
Maybe it might have been useful if you had done it using the product rule.

8. Re: dQ/dt

$\displaystyle \frac{(1+bt^2)^2[(1+bt^2)(a)-(at)(3)(2bt)]}{(1+bt^2)^6}$
$\displaystyle \frac{a+abt^2-6abt^2}{(1+bt^2)^4}$
$\displaystyle \frac{a(1-5bt^2)}{(1+bt^2)^4}$

Thank you all for helping. The above should be the solved equation I think.