simplification of a derivated function...

[thepassenger48]

I have no clue to get from step 1 to step 2, and then from step 2 to step 3.
I would be very grateful if someone could explain to me what is hapening. I have a test tomorrow and I'm on the verge of panicking!

Thank you very much.

 

[soroban]

Hello, thepassenger48!

It's just algebra . . .


We are given:\(\displaystyle \L\:f(x) \;= \;\frac{6x^2}{\left(x^3\,+\,1\right)^2}\)

Then: \(\displaystyle \L\:f'(x)\;=\;\frac{\left(x^3\,+\,1\right)^2\cdot12x\,-\,6x^2\cdot2\left(x^3\,+\,1\right)\cdot 3x^2}{\left[\left(x^3\,+\,1\right)^2\right]^2}\)

Simplify: \(\displaystyle \L\:f'(x) \;= \;\frac{12x\left(x^3\,+\,1\right)^2\,-\,36x^4\left(x^3\,+\,1\right)}{\left(x^3\,+\,1\right)^4}\)

Factor: \(\displaystyle \L\:f'(x) \;= \;\frac{12x\left(x^3\,+\,1\right)\,\cdot\,\left[\left(x^3\,+\,1\right)\,-\,3x^3\right]}{\left(x^3\,+\,1\right)^4}\)

Reduce: \(\displaystyle \L\:f'(x)\;=\;\frac{12x\left(x^3\,+\,1\,-\,3x^3\right)}{(x^3\,+\,1)^3}\)

Therefore: \(\displaystyle \L\:f'(x)\;=\;\frac{12x\left(1\,-\,2x^3\right)}{\left(x^3\,+\,1\right)^3}\)
.

 

[thepassenger48]

Ahh! It's the factoring part that had me confused! Thank you very much soroban.
I guess that's what happens when you don't take any math classes for 3 years, my algebra skills are very rusty! hehe