JAMB 1999 · UME · Q37

If $y = 3x^2(x^3 + 1)^{\frac12}$ , find $\frac{dy}{dx}$ .

Worked solution (try it first)

Product rule with $u = 3x^2$ and $v = (x^3 + 1)^{1/2}$ .
Then $u' = 6x$ and, by the chain rule, $v' = \frac12(x^3 + 1)^{-1/2} \times 3x^2$
$= \dfrac{3x^2}{2(x^3 + 1)^{1/2}}$ .
So $\frac{dy}{dx} = 6x(x^3 + 1)^{1/2} + 3x^2 \times \dfrac{3x^2}{2(x^3 + 1)^{1/2}}$ .
Over the common denominator $2(x^3 + 1)^{1/2}$ : $\dfrac{12x(x^3 + 1) + 9x^4}{2(x^3 + 1)^{1/2}}$ , option D.