# I cannot find the derivatives of these functions using the product rule?

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.

1. ${(3x^2)(3x+7)^2}+{(x^36)(3x+7)}$
2. $5(2x+1)^4 (2) (3x+2)^4 + (2x+1)^54(3x+2)^3(3)$
• You need both chain rule and product rule. Commented Feb 21, 2014 at 23:54
• @user60887 The OP says the teacher forbade use of the chain rule. In any case, it is not necessary, although it is more convenient. Commented Feb 21, 2014 at 23:55
• Shouldn't 1) be $(3x^2)(3x+7)^2+(x^36)(3x+7)$
– user122283
Commented Feb 21, 2014 at 23:56
• @SanathDevalapurkar Sorry, that is the right answer. Commented Feb 21, 2014 at 23:59
• 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. Commented Feb 22, 2014 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:

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

• Thank you so much!! I couldn't get this question and it was really bugging me! Commented Feb 22, 2014 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.

• The OP says that the teacher forbade use of the chain rule. Commented Feb 21, 2014 at 23:56
• @Arkamis I changed my answer, thanks.
– user122283
Commented Feb 21, 2014 at 23:57
• Thank you for your help!! I really appreciate it!! Commented Feb 22, 2014 at 0:02
• @user130599 Sure - anything for a fellow mathematician!
– user122283
Commented Feb 22, 2014 at 0:03

The easiest solution (by this I mean no fancy theorems needed) is to calculate the product. So we have to way to do so. The first way: \begin{align} y&=x^3(3x+7)^2\\&=x^3(9x^2+42x+49), \\ \Longrightarrow y'&=3x^2(9x^2+42x+49)+x^3(18x+42)\\&=45x^4+168x^3+147x^2, \end{align}

The second way: \begin{align} y&=x^3(3x+7)^2\\&=x^3(9x^2+42x+49)\\&=9x^5+42x^4+49x^3, \\\Longrightarrow y'&=45x^4+168x^3+147x^2. \end{align}

From this point, both ways continue as \begin{align} y'&=(x^2)\cdot(45x^2+168x+147)\\ &=(4)\cdot (45\cdot 4-168\cdot 2+147)\\ &=4\cdot(180-336+147)\\ &=4\cdot(327-336)\\ &=-36\end{align}

Note that for the other exercise you can use the following formulas from college algebra, $$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4,\\ (x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+x^5.$$

So,

$$(2x+1)^5=32x^5+80x^4+80x^3+40x^2+10x+1,\\ (3x+2)^4=81x^4+216x^3+216x^2+96x+16.$$