Problem with simplifying fractions

[jcheema]

Problem with factorising

Hi, the question is to differentiate the following equation with respect to x.

x^4(3x-1)^3

Using the product rule i think i'v partially completed this to

x^4(3(3x-1)^2) + (3x-1)^3(4x^3)

I'm now required to simplify this - which leaves me completely stumped. There is an answer in the back of the book which i simply cannot get to, could i please hear some of your solutions to this!

James

Edit:Wrong title

 

[danago]

There is a slight error in that derivative you have come to.

<br /> \frac{d}{{dx}}((3x - 1)^3 ) = 9(3x - 1)^2<br /> <br />

You seem to have put 3 instead of 9

Other than that, it looks fine to me. After changing it, try taking out a factor of <br /> x^3 (3x - 1)^2 <br />

 

[jcheema]

So it would now become

x^4(9(3x-1)^2) + (3x-1)^3(4x^3) - i see my mistake there, my only problem now is how to simplify this further. How would i go about taking the factor out?

 

[danago]

I replied to your private message explaining everything. If the equations i wrote in that don't show up, ill post it here :)

 

[danago]

Well it seems that the LaTeX wouldn't work in private messages. Here is exacly what i messages you:

Hi there danago, thankyou for your reply to the problem i was having earlier. Although I'm still struggling in how you solve this kind of equation by factorising and simplifying, if you have the time would you please explain in simpler terms how to work out this equation. Thanks for your time, James

Sure. You had the equation:

<br /> f(x) = x^4 (3x - 1)^3 <br />

Your differentiation of that was almost perfect, but you made one slight error. Break that function up into two separate functions, x^4
and (3x - 1)^3. You got the derivative of the first one right, but the second one was slightly off. Use the chain rule to differentiate it.

The chain rule states that:
<br /> \frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}<br />

Therefore, if we let:
<br /> \begin{array}{l}<br /> u = 3x - 1 \\ <br /> y = u^3 \\ <br /> \end{array}<br />

We can then differentiate it:
<br /> \begin{array}{l}<br /> \frac{{dy}}{{du}} = 3u^2 \\ <br /> \frac{{du}}{{dx}} = 3 \\ <br /> \therefore \frac{{dy}}{{dx}} = 9u^2 = 9(3x - 1)^2 \\ <br /> \end{array}<br />

Therefore, from the product rule, we get:

<br /> \frac{d}{{dx}}(x^4 (3x - 1)^3 ) = 9x^4 (3x - 1)^2 + 4x^3 (3x - 1)^3 <br />

Now, you could leave it like that, but its a bit messy. So you now need to factorize it. If you had something like this:

<br /> y = ab + ac<br />

You could easily factorize by taking out a factor of a.

<br /> y = ab + ac = a(b + c)<br />

The same principal applies to the derivative. If you find it a little hard to do, try this:

Let:
<br /> \begin{array}{l}<br /> a = 3x - 1 \\ <br /> \therefore \frac{{dy}}{{dx}} = 9x^4 a^2 + 4x^3 a^3 \\ <br /> \end{array}<br />

Now notice that each term has an 'a' and an 'x' in it. Just factorize them both out, like so:

<br /> 9x^4 a^2 + 4x^3 a^3 = x^3 a^2 (9x + 4a)<br />

And then substitute a=3x-1 back into that, to get:
<br /> \begin{array}{c}<br /> x^3 a^2 (9x + 4a) = x^3 (3x - 1)^2 (9x + 4(3x - 1)) \\ <br /> = x^3 (3x - 1)^2 (21x - 4) \\ <br />

And that's it. I hope that helped. If you didnt understand any of what i said, feel free to message back. I am more than willing to help.

Dan.

PS. I am not sure if TeX works in private messages. If my equations arent working, just tell me, and ill fix them.

 

[jcheema]

Excellent description there :) Beats my textbook! However I'm now just struggling to see how

<br /> 9x^4 a^2 + 4x^3 a^3 = x^3 a^2 (9x + 4a)<br />

As i'd work out x^3*9x, which would give you the 9x^4. But I'm not really sure how it'd then multiply out to give you the other terms, apologies as my maths skills are not the best and i ask you to please have patience with me!

 

[danago]

ok. When you are factorizing things, essentially, you are dividing through by something, and then multiplying by that same thing again.

Look at:
<br /> 9x^4 a^2 + 4x^3 a^3 <br />

You can divide the whole expression by x, to get:

<br /> 9x^3 a^2 + 4x^2 a^3 <br />

But you can't just divide like that, because you are then changing the value of the expression. So to "cancel out" this division process, you then do the opposite of that division by x, which would be to multiply by x.

<br /> x(9x^3 a^2 + 4x^2 a^3 )<br />

What about now if i divided by 'xa'.

<br /> 9x^3 a + 4x^2 a^2<br />

To make that division process valid, i must now multiply again by 'xa', and i get:

<br /> xa(9x^3 a + 4x^2 a^2)<br />

What about if we do the same now, but instead divide by x^3 a^2.

<br /> \frac{{9x^4 a^2 + 4x^3 a^3 }}{{x^3 a^2 }} = 9x + 4a<br />

But dividing by that completely changed the value of the expression. So to make that division valid, we now have to multiply the whole thing by what we divided it by:

<br /> x^3 a^2 (9x + 4a)<br />

And that's where the 9x^4 a^2 + 4x^3 a^3 = x^3 a^2 (9x + 4a) came from.

Did that make any sense to you?

 

[jcheema]

That makes perfect sense yes, thanks very much for your time - you're a credit to the forums

 

[danago]

No problems :) Glad to be of assistance :)

Im glad you understand it now :)