Solving it would give $$\frac{(2x-3)(2x)-(x^2+1)(2)}{(2x+3)^2}$$ $$=\frac{2x^2-6x-2}{(2x+3)^2}$$ I don’t think it’s possible to move ahead from here. The answer is $$\frac{6x-2}{(2x+3)^2}$$ I have serious doubts that the answer given is wrong, but I would like to verify it here nonetheless. Thanks a lot as always.
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2wrong sign on the first term of the numerator: should be $2x + 3$ and not $2x - 3$ – D. Ungaretti Aug 13 '19 at 16:13
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Yes, I was being dumb. Thanks for the help. – Aditya Aug 13 '19 at 16:15
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1Your recent edit made my answer pointless. Please refrain from editing a question once an answer is posted. – Shubham Johri Aug 13 '19 at 16:20
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Of course. It was a typo, so I figured I would correct it. I will cancel the edit if you want it. – Aditya Aug 13 '19 at 16:22
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I have already done that – Shubham Johri Aug 13 '19 at 16:25
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Yeah, no problem. – Aditya Aug 13 '19 at 16:25
1 Answers
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$$y'=\frac{(2x\color{red}+3)(2x)-(x^2+1)(2)}{(2x+3)^2}=\frac{2(x^2+3x-1)}{(2x+3)^2}$$
Shubham Johri
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