Differentiation of 1/3x(4-x)^2

[cr7]

Ok well i've been tryin to derive this equation for a while.
I've tried using chain rule and then tried product rule, but im not sure if im doing it right.

y= (1/3x)(4-x)^2
y=(3x^-1)(8-2x)
y'=(8-2x)(-3x^-2)+ (-3x^-1)(-2)
y'= (-15x^-2 +6x^-1) + (6x^-1)
y'= -15x^-2 + 12x^-1
y'= -1/15x^2 +1/12x

If someone could help it would be greatly appreciated.
Thank you

 

[galactus]

Is that:

\(\displaystyle \L\\\frac{1}{3x(4-x)^{2}}\).

To check, integrate your result. Do you get the original expression or function?.

 

[cr7]

actually the equation is:
y= (1/3x) (4-x)^2
sorry for the confusion

 

[skeeter]

one more bit of confusion, please clarify ...

is it \(\displaystyle \L \left(\frac{1}{3} x\right) (4-x)^2\)

or \(\displaystyle \L \L \left(\frac{1}{3x}\right) (4-x)^2\)

?

 

[cr7]

one more bit of confusion, please clarify ...

is it \(\displaystyle \L \left(\frac{1}{3} x\right) (4-x)^2\)

or \(\displaystyle \L \L \left(\frac{1}{3x}\right) (4-x)^2\)

?

It's \(\displaystyle \L \left(\frac{1}{3} x\right) (4-x)^2\)

 

[skeeter]

\(\displaystyle \L \frac{x}{3} (4-x)^2\)

product and chain rule ...

\(\displaystyle \L \frac{x}{3} \cdot 2(4-x) \cdot (-1) + (4-x)^2 \cdot \frac{1}{3}\)

\(\displaystyle \L \frac{4-x}{3} [-2x + (4-x)]\)

\(\displaystyle \L \frac{(4-x)(4-3x)}{3}\)

 

[cr7]

Ok thank you soo much