Use the product, power, and chain rules.
![y = (x^2 + 1) (x^3 + 1)^3](https://img.qammunity.org/2022/formulas/mathematics/high-school/phb9kd5q9u8x0cb5y1a4a0sj98mdot65md.png)
Differentiate both sides:
![(\mathrm dy)/(\mathrm dx) = (\mathrm d\left((x^2 + 1) (x^3 + 1)^3\right))/(\mathrm dx)](https://img.qammunity.org/2022/formulas/mathematics/high-school/yqrzgzrl8u7kaau1rd8a5kjybuu2pllh9u.png)
Product rule:
![(\mathrm dy)/(\mathrm dx) = (\mathrm d(x^2 + 1))/(\mathrm dx)(x^3+1)^3 + (x^2+1)(\mathrm d(x^3 + 1)^3)/(\mathrm dx)](https://img.qammunity.org/2022/formulas/mathematics/high-school/clb9wxedegdx5fyjt9x73v1ot28nb1lrsm.png)
Power rule for the first derivative, power and chain rules for the second one:
![(\mathrm dy)/(\mathrm dx) = 2x(x^3+1)^3 + 3(x^2+1)(x^3+1)^2(\mathrm d(x^3 + 1))/(\mathrm dx)](https://img.qammunity.org/2022/formulas/mathematics/high-school/bep7kzp8xf8xpq7e3kbfj3rn1rcor30p4i.png)
One last applicaton of power rule:
![(\mathrm dy)/(\mathrm dx) = 2x(x^3+1)^3 + 9x^2(x^2+1)(x^3+1)^2](https://img.qammunity.org/2022/formulas/mathematics/high-school/ok717pprap4h1af303ky2cxw4lsmkp0nn9.png)
You could stop here, or continue and simplify the result by factorizing:
![(\mathrm dy)/(\mathrm dx) = x(x^3+1)^2 \left(2(x^3+1) + 9x(x^2+1)\right) \\\\ (\mathrm dy)/(\mathrm dx) = \boxed{x(x^3+1)^2 (11x^3+9x+2)}](https://img.qammunity.org/2022/formulas/mathematics/high-school/hmwj4ikakkplw258g74baoxbwm6pjzwcjs.png)