I used the power rule and the chain rule and ended up with this: $$y'= (3x-1)^2 \times 2(2x+3) \times 2 + (2x+3)^2 \times 2(3x-1)\times 3$$

The next step, which I do not understand how it is combined or created is this:

$$y'= 2(3x-1)(2x+3)\left[2\cdot(3x-1)+3(2x+3)\right]$$ Where did the exponents go? What is combined? How is it combined?

The Final Answer should be this, according to my teacher: $$y'= 2(3x-1)(2x+3)(12x+18)$$

  • $\begingroup$ No it's all correct. $\endgroup$
    – Britney
    Sep 30, 2014 at 6:44
  • 1
    $\begingroup$ @Nick Assuming the OP copied the expected answer correctly, your edit is fine; it's just that the teacher's answer is incorrect. (The $18$ should be a $7$.) Three of us + Maple (I checked) have come up with the correct answer. :-) $\endgroup$
    – James
    Sep 30, 2014 at 6:51

2 Answers 2


I think that you forgot that the derivative of $u^n(x)=n u^{n-1}(x)u'(x)$.

I suppose that you want the derivative of $$y=(3x-1)^2 (2x+3)^2$$ So, let us set $u=(3x-1)^2$ and $v=(2x+3)^2$. So $y=u\times v$ and $y'=u'\times v+u\times v'$.

Now $$u'=2\times(3x-1)'\times(3x-1)=6(3x-1)$$ $$v'=2\times(2x+3)'\times(2x+3)=4(2x+3)$$ $$u'\times v+u\times v'=6(3x-1)(2x+3)^2+4(2x+3)(3x-1)^2$$ Factor $(3x-1)(2x+3)$; so $$u'\times v+u\times v'=(3x-1)(2x+3)\Big(6(2x+3)+4(3x-1)\Big)$$ Expand what is inside the last bracket and simplify to arrive to $$y'=2 (2 x+3) (3 x-1) (12 x+7)$$

With reference to the exponents, look at the first line and think about $n=2$ and remember that we never use exponent $1$ in such context of polynomials.

  • $\begingroup$ Oh! Thank you so much, I didn't realize she factored it out. And yes, my math teacher makes a lot of mistakes so it wouldn't be unusual for the answer to be different. $\endgroup$
    – Britney
    Sep 30, 2014 at 6:53
  • $\begingroup$ You are very welcome. When you have powers such as in $y=u^n \times v^m$, the common factor will be $u^{n-1}\times v^{m-1}$. Just for fun, prove it. Cheers :-) $\endgroup$ Sep 30, 2014 at 6:58




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