Question

Find the derivatives of $y=(x^2+1)(x^3+3)$.

Answer 1

From the product rule with $u=x^2+1$ and $v=x^3$, we find
$\frac{d}{dx}\left[\right(x^2+1\left)\right(x^3+3\left)\right]=(x^2+1)\frac{d}{dx}(x^3+3)+(x^3+3)\frac{d}{dx}(x^2+1)$
$(x^2+1)\left(3x^2\right)+(x^3+3)\left(2x\right)$
$=3x^4+3x^2+2x^4+6x$
$=5x^4+3x^2+6x$.

The above sum can be done as well (perhaps better multiplying out the original expression for $y$ and differential the resulting polynomial. We now check;

$y=(x^2+1)(x^3+3)=x^5+x^3+3x^2+3$

$\frac{dy}{dx}=5x^4+3x^2+6x$

This is in agreement with our first calculation.