# Thread: homework help needed with differentiation

1. ## homework help needed with differentiation

I have attached a question which i kind of confused with. I understand the differentiation parts but i don't understand the simplifying bits.

The main simplification bit i don't understand is the 12x^2 -4x bit which i have circled in the attached image. I think that the 12x^2 -4x bit comes from the simplifying (squareroot 3x-1) * 2x but i not sure and if it does i don't understand this simplification on how it was done.

I would greatly appreichate any help in solving this problem.

thanks

2. Originally Posted by rpatel
I have attached a question which i kind of confused with. I understand the differentiation parts but i don't understand the simplifying bits.

The main simplification bit i don't understand is the 12x^2 -4x bit which i have circled in the attached image. I think that the 12x^2 -4x bit comes from the simplifying (squareroot 3x-1) * 2x but i not sure and if it does i don't understand this simplification on how it was done.

I would greatly appreichate any help in solving this problem.

thanks
we have $f'(x) = x^2 \cdot \frac 32 (3x - 1)^{- \frac 12} + 2x \sqrt {3x - 1}$

$= \frac {3x^2}{2 \sqrt {3x - 1}} + 2x \sqrt {3x - 1}$

the LCD here is $2 \sqrt {3x - 1}$, so we get:

$f'(x) = \frac {3x^2 + 2 \cdot 2x \left( \sqrt {3x - 1} \right)^2}{2 \sqrt {3x - 1}}$

$= \frac {3x^2 + 4x(3x - 1)}{2 \sqrt {3x - 1}}$

$= \frac {3x^2 + 12x^2 - 4x}{2 \sqrt {3x - 1}}$

...

3. Originally Posted by rpatel
Let's set $y=f(x)=x^2\sqrt{3x-1}\implies y^2=x^4(3x-1)=3x^5-x^4$

Let's take now the derivative

$2yy'=15x^4-4x^3\implies y'=\frac{x^3(15x-4)}{2y}$

Back substitute and we happily get that $f'(x)=\frac{x^3(15x-4)}{2x^2\sqrt{3x-1}}=\frac{x(15x-4)}{2\sqrt{3x-1}}$

Does that make sense?