<s>[INST] Some chain rule questions. Answer checks I have these problems and I just wanted to make sure I was doing Chain Rule correctly and that no further simplification was possible:


*

*$$y = (2x^3 + 5)^4$$
$$ so \frac{dy}{dx} = 4(2x^3+5)^3 * 6x^2$$


$$ = 24x^2(2x^3 + 5)^3$$


*$$f(x) = (5x^6 + 2x^3)^4$$
$$f'(x) = 4(5x^6 + 2x^3)^3 *(30x^5 + 6x^2)$$
3.
$$f(x) = (1 + x + x^2)^{99}$$
$$f'(x) = 99(1 + x + x^2)^{98} * (1 + 2x)$$
4.
$$f(x) = \sqrt{5x + 1}$$
$$f'(x) = \frac{1}{2} * (5x+1)^{-\frac{1}{2}} * 5$$
$$\frac{2.5}{\sqrt{5x + 1}}$$


*I could use some help on this one:


$$f(x) = (2x-3)^4 * (x^2+x+1)^5$$
I started with this:
$$ f'(x) = (2x-3)^4 * 5(x^2+x+1)^4 * (2x+1) + (x^2 +x+1)^5 * 4(2x-3)^3 * 2$$
But how do I simplify from here?
 [/INST] My comment was getting too long so I'll post a short answer.
For number $5$, as mentioned in the comments you should use the product rule. Recall that:
$$\frac{d}{dx}f(x)g(x)=\frac{df(x)}{dx}g(x)+f(x)\frac{dg(x)}{dx}.$$
Now, set $$\begin{aligned}&f(x)=(2x-3)^4\\&g(x)=(x^2+x+1)^5\end{aligned}$$
and differentiate both (using the chain rule as you did on the previous exercises)
$$\begin{aligned}&\frac{df(x)}{dx}=8(2x-3)^3\\&\frac{dg(x)}{dx}=5(2x+1)(x^2+x+1)^4\end{aligned}$$
so
$$\frac{d}{dx}\left((2x-3)^4(x^2+x+1)^5\right)=8(2x-3)^3(x^2+x+1)^5+5(2x+1)(2x-3)^4(x^2+x+1)^4$$
and by simplifying furthermore, you obtain
$$(2 x-3)^3 \left(x^2+x+1\right)^4 \left(28 x^2-12 x-7\right).$$
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