Find the derivative of $y = (x^{2} + 1) (x^{3} + 3)$ .

5 x^{4} + 3 x^{2} + x

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4 x^{4} + 3 x^{2} + 6 x

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5 x^{4} + 2 x^{2} + 6 x

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5 x^{4} + 3 x^{2} + 6 x

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Solution

The correct option is D $5 x^{4} + 3 x^{2} + 6 x$
From the product rule with $u = x^{2} + 1$ and $v = x^{3} + 3$ , we find
$\frac{d}{d x} [(x^{2} + 1) (x^{3} + 3)] = (x^{2} + 1) (3 x^{2}) + (x^{3} + 3) (2 x) = 3 x^{4} + 3 x^{2} + 2 x^{4} + 6 x \frac{d y}{d x} = 5 x^{4} + 3 x^{2} + 6 x$

Alternative:
$y = (x^{2} + 1) (x^{3} + 3) = x^{5} + x^{3} + 3 x^{2} + 3 \frac{d y}{d x} = 5 x^{4} + 3 x^{2} + 6 x$

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