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

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

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

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

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

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Solution

The correct option is A $5 x^{4} + 3 x^{2} + 6 x$
From the product Rule with $u = x^{2} + 1$ and $v = x^{3} + 3$
We find, $\frac{d}{d x} [(x^{2} + 1) (x^{3} + 1)] = (x^{2} + 1) (3 x^{2}) + (x^{3} + 3) (2 x)$
$= 3 x^{4} + 3 x^{2} + 2 x^{4} + 6 x$
$= 5 x^{4} + 3 x^{2} + 6 x$
Alternatively, it can be done as well (perhaps better way) by multiplying the original expression for y and differentiating the resulting polynomial.
$y = (x^{2} + 1) (x^{3} + 1) = x^{5} + x^{3} + 3 x^{2} + 3$
$\frac{d y}{d x} = 5 x^{4} + 3 x^{2} + 6 x$
This is in agreement with our first calculation.

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