Differentiate with product rule Question: differentiate $x(x^2 +3x)^3$
I have gotten to the point where i've used the product rule and i've gotten
$$(x^2 + 3x)^3 + x\cdot(3x+9)(x^2 + 3x)^2$$
but now that it comes to the simplifying im completely at a loss, any help would greatly appreciated
 A: Using the product rule is not enough. 
$$(x(x^2 + 3x)^3)^\prime = (x)^\prime \cdot (x^2 + 3x)^3 + x \cdot ((x^2 + 3x)^3)^\prime$$
$$=  (x^2 + 3x)^3 + x \cdot ((x^2 + 3x)^3)^\prime$$
For the rigth part you need to use the chain rule. 
$$=  (x^2 + 3x)^3 + x \cdot 3(x^2 + 3x)^2\cdot(x^2+3x)^\prime$$
$$=  (x^2 + 3x)^3 + x \cdot 3(x^2 + 3x)^2\cdot(2x+3)$$
Now you can simplifying the result:
$$=  (x^2 + 3x)^2 \cdot \left((x^2 + 3x) + x \cdot 3\cdot(2x+3)\right)$$
$$=  (x^2 + 3x)^2 \cdot \left(x^2 + 3x + 6x^2 + 9x\right)$$
$$=  (x^2 + 3x)^2 \cdot \left(7x^2 + 12x\right)$$
A: Note that the product rule is defined as
\[
\frac{d}{dx}[f(x)\cdot g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]
\]
And the chain rule is defined as
\[
f(x)=g(h(x))
\]
\[
\frac{d}{dx}[f(x)]=\frac{d}{dx}[g(h(x))] \frac{d}{dx}[h(x)]
\]
So now here are the steps 
\[
\frac{d}{dx}\left[x(x^{2}+3x)^{3}\right]=\frac{d}{dx}\left[x^{4}(x+3)^{3}\right] =x^{4}\frac{d}{dx}\left[(x+3)^{3}\right]+(x+3)^{3}\frac{d}{dx}\left[x^{4}\right]
\]
\[
=3x^{4}(x+3)^{2}\frac{d}{dx}[x+3]+4x^{3}(x+3)^{3}=3x^{4}(x+3)^{2}+4x^{3}(x+3)^{3}
\]
\[
=x^{3}(x+3)^{2}(3x+4x+12)=x^{3}(x+3)^{2}(7x+12)
\]
A: Use the $\color{red}{\text{product rule}}$ and the $\color{blue}{\text{chain rule}}$
$$\begin{array}{rcl}(x(x^2+3x)^3)' &\color{red}{=}& x'\cdot (x^2+3x)^3+((x^2+3x)^3)'\cdot x \\ &\color{blue}{=}& (x^2+3x)^3+x(3(x^2+3x)^2\cdot (x^2+3x)') \\ &=& (x^2+3x)^3+x(3(x^2+3x)^2(2x+3))\\ &=& (x^2+3x)^3+3x(x^2+3x)^2(2x+3)\\ &=&x^3 (x+3)^2 (7 x+12) \end{array}$$ 
