<s>[INST] Where did I go wrong with my differentiation here? This question involves the product rule and power of a function rule for differentiation. I often make silly algebra errors, but here I cannot find them. The answer is supposed to be y'=22. Thanks!
When $x=-1$, $y=(2x+1)^5(3x+2)^4$.
Hence, the derivative,
$$y'=5(2x+1)^4 \cdot (2)(3x+2)^4 + (2x+1)^5 \cdot 4(3x+2)^3 \cdot 3$$
Plugging in $x=-1$,
$$y'= 5(2(-1)+1)^4 \cdot (2)(3(-1)+2)^4 + (2(-1)+1)^5 \cdot 4(3x+2)^3 \cdot 3\\
y' = 10-12\\
y' = -2$$
 [/INST] We are computing the derivative at $x=-1$ when $y=(2x+1)^5 (3x+2)^4$. To find the derivative, we will be using chain rule and the product rule in correlation.
The product rule: $f'g+fg'$
In our case, Let $f=(2x+1)^5$ and $g=(3x+2)^4$
Hence, $f'=5(2x+1)^{5-1}\cdot(2x+1)^{'}\\ \;\;\;=5(2x+1)^4\cdot 2\\ \;\;\;=10(2x+1)^4$ and $g'=4(3x+2)^{4-1}\cdot (3x+2)^{'}\\ \;\;\;=4(3x+2)^3\cdot3\\ \;\;\;=12(3x+2)^3$
Completing the product rule ($f'g+fg'$): $$y'=10(2x+1)^4*(3x+2)^4+(2x+1)^5*12(3x+2)^3$$
Evaluating at $x=-1$, $$y'(-1)=10(2(-1)+1)^4\cdot (3(-1)+2)^4 +(2(-1)+1)^5\cdot 12(3(-1)+2)^3\\y'(-1) =10+12
\\y'(-1)=22$$
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