Use the product, power, and chain rules.
[tex]y = (x^2 + 1) (x^3 + 1)^3[/tex]
Differentiate both sides:
[tex]\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d\left((x^2 + 1) (x^3 + 1)^3\right)}{\mathrm dx}[/tex]
Product rule:
[tex]\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d(x^2 + 1)}{\mathrm dx}(x^3+1)^3 + (x^2+1)\dfrac{\mathrm d(x^3 + 1)^3}{\mathrm dx}[/tex]
Power rule for the first derivative, power and chain rules for the second one:
[tex]\dfrac{\mathrm dy}{\mathrm dx} = 2x(x^3+1)^3 + 3(x^2+1)(x^3+1)^2\dfrac{\mathrm d(x^3 + 1)}{\mathrm dx}[/tex]
One last applicaton of power rule:
[tex]\dfrac{\mathrm dy}{\mathrm dx} = 2x(x^3+1)^3 + 9x^2(x^2+1)(x^3+1)^2[/tex]
You could stop here, or continue and simplify the result by factorizing:
[tex]\dfrac{\mathrm dy}{\mathrm dx} = x(x^3+1)^2 \left(2(x^3+1) + 9x(x^2+1)\right) \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \boxed{x(x^3+1)^2 (11x^3+9x+2)}[/tex]
