А затем найти вторую производную.
y`=2·(1/2)·(4x+3)–1/2·(4x+3)` – 3·(–1/2)·(x3+x+1)–3/2·(x3+x+1)`;
y`=4·(4x+3)–1/2+(3/2)·(3x2+1)·(x3+x+1)–3/2
y``=4·(–1/2)·(4x+3)–3/2·4 +(3/2)·(3x2+1)`·(x3+x+1)–3/2+(3/2)·(3x2+1)·(–3/2)·(x3+x+1)–5/2·(x3+x+1)`
y``=–8·(4x+3)–3/2+9x2·(x3+x+1)–3/2+(9/4)(3x2+1)2·(x3+x+1)–5/2