I need help finding the derivative of $y=x^3(3x+7)^2$ at $x=-2$ I tried to simplify the function to $y=x^3 (3x+7)(3x+7)$ and the simplify it into two terms and the derivatives of those terms using the product rule, but that doesn't work. Since I don't understand how to approach this type of question (with the power on the outside), I couldn't do this one either: $y=(2x+1)^5(3x+2)^4$ at $x=-1$

I don't need help finding the derivative at the specific values of $x$, I just can't seem to get the general case. BTW, I can only use the product rule as my teacher did not teach the quotient and chain rule to us, so she doesn't allow us to use them.

the answers(aka the derivative):

  1. ${(3x^2)(3x+7)^2}+{(x^36)(3x+7)}$
  2. $5(2x+1)^4 (2) (3x+2)^4 + (2x+1)^54(3x+2)^3(3)$
  • $\begingroup$ You need both chain rule and product rule. $\endgroup$ – user60887 Feb 21 '14 at 23:54
  • $\begingroup$ @user60887 The OP says the teacher forbade use of the chain rule. In any case, it is not necessary, although it is more convenient. $\endgroup$ – Emily Feb 21 '14 at 23:55
  • $\begingroup$ Shouldn't 1) be $(3x^2)(3x+7)^2+(x^36)(3x+7)$ $\endgroup$ – user122283 Feb 21 '14 at 23:56
  • $\begingroup$ @SanathDevalapurkar Sorry, that is the right answer. $\endgroup$ – user130599 Feb 21 '14 at 23:59
  • $\begingroup$ Has your teacher showed you that the derivative of $(ax+b)^n$ is $an(ax+b)^{n-1}$? Maybe she will allow you to use this, even though in fact it is a special case of the chain rule. If this is forbidden then your second question is a product of nine terms, which can be done but is going to take ages. Alternatively you could multiply out by using the binomial theorem, $(2x+1)^5=32x^5+\{{\rm etc}\}$, but this will also be a lot of work. $\endgroup$ – David Feb 22 '14 at 0:00

The product rule is: if $f(x) = g(x)h(x)$, then $f'(x) = g'(x)h(x)+g(x)h'(x)$. Extending it to three terms, we get:

$$f(x) = g(x)h(x)j(x) \implies f'(x) = g'(x)\left[h(x)j(x)\right]+g(x)\left[h(x)j(x)\right]' \\ = g'(x)h(x)j(x)+g(x)\left[h'(x)j(x)+h(x)j'(x)\right] \\ = g'(x)h(x)j(x)+g(x)h'(x)j'(x)$$ as we might expect.

So, solving via the product rule, $$f'(x) = (x^3)'(3x+7)(3x+7)+x^3(3x+7)'(3x+7)+x^3(3x+7)(3x+7)'.$$

Notice that the last two terms are the same.

$$f'(x) = 3x^2(3x+7)^2+2x^3(3x+7)'(3x+7) \\ = 3x^2(3x+7)^2+2x^3(3x+7)(3) \\ = 9x^3+21x^2+3(6x^4+14x^3) \\ = 18x^4+51x^3+21x^2.$$

This is not the easiest way to solve the problem. The easiest way is to use the product rule once, and the chain rule once:

$$f'(x) = 3x^2(3x+7)^2+x^3\cdot 2 \cdot(3x+7)\cdot (3x+7)' \\ = 9x^3+21x^2+2x^3(3x+7)(3) \\ = 18x^4+51x^3+21x^2.$$

  • $\begingroup$ Thank you so much!! I couldn't get this question and it was really bugging me! $\endgroup$ – user130599 Feb 22 '14 at 0:01

So, $y=x^3(3x+7)^2$. Thus, the product rule $$\frac{d(uv)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$$ Gives $$y^{'}=x^3\frac{d(3x+7)^2}{dx}+(3x+7)^2(3x^2)$$ You can use easily calculate $\frac{d(3x+7)^2}{dx}$: $$\frac{d(3x+7)^2}{dx}=2(3x+7)(3)$$ You should now be able to solve all your other problems using this method.

  • $\begingroup$ The OP says that the teacher forbade use of the chain rule. $\endgroup$ – Emily Feb 21 '14 at 23:56
  • $\begingroup$ @Arkamis I changed my answer, thanks. $\endgroup$ – user122283 Feb 21 '14 at 23:57
  • $\begingroup$ Thank you for your help!! I really appreciate it!! $\endgroup$ – user130599 Feb 22 '14 at 0:02
  • $\begingroup$ @user130599 Sure - anything for a fellow mathematician! $\endgroup$ – user122283 Feb 22 '14 at 0:03

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