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.
the answers(aka the derivative):


*

*${(3x^2)(3x+7)^2}+{(x^36)(3x+7)}$

*$5(2x+1)^4 (2) (3x+2)^4 + (2x+1)^54(3x+2)^3(3)$

 A: 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:
$$f'(x) = 3x^2(3x+7)^2+x^3\cdot 2 \cdot(3x+7)\cdot (3x+7)' \\
= 9x^3+21x^2+2x^3(3x+7)(3) \\
= 18x^4+51x^3+21x^2.$$
A: 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.
A: 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,
$$(2x+1)^5=32x^5+80x^4+80x^3+40x^2+10x+1,\\
(3x+2)^4=81x^4+216x^3+216x^2+96x+16.$$
