Question

find the derivative of $f(x)=6x^{2}sqrt{2x^{2}-1}$. select the correct answer below:
$12x(x^{2}+1)$
$\frac{12x(3x^{2}-1)}{sqrt{2x^{2}-1}}$
$12x(2x^{2}-1)(x^{2}+1)$
$\frac{12x(3x^{2}-1)}{sqrt{x^{2}+2x - 1}}$

Explanation:

Step1: Apply product - rule

$(uv)^\prime = u^\prime v+uv^\prime$, where $u = 6x^{2}$ and $v=\sqrt{2x^{2}-1}=(2x^{2}-1)^{\frac{1}{2}}$.
$u^\prime=12x$, $v^\prime=\frac{1}{2}(2x^{2}-1)^{-\frac{1}{2}}\cdot4x=\frac{2x}{\sqrt{2x^{2}-1}}$.

Step2: Calculate $f^\prime(x)$

$f^\prime(x)=12x\sqrt{2x^{2}-1}+6x^{2}\cdot\frac{2x}{\sqrt{2x^{2}-1}}=\frac{12x(2x^{2}-1)+12x^{3}}{\sqrt{2x^{2}-1}}=\frac{24x^{3}-12x + 12x^{3}}{\sqrt{2x^{2}-1}}=\frac{12x(3x^{2}-1)}{\sqrt{2x^{2}-1}}$.

Answer:

$\frac{12x(3x^{2}-1)}{\sqrt{2x^{2}-1}}$ (corresponding to the second option in the multiple - choice list)