attempt 1: 10 attempts remaining. find the derivative. y = (√x + 2)(x² - 1); y=

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u=\sqrt{x}+2=x^{\frac{1}{2}}+2$ and $v = x^{2}-1$.

Step2: Find $u'$ and $v'$

Differentiate $u$ with respect to $x$: $u'=\frac{d}{dx}(x^{\frac{1}{2}}+2)=\frac{1}{2}x^{-\frac{1}{2}}$. Differentiate $v$ with respect to $x$: $v'=\frac{d}{dx}(x^{2}-1)=2x$.

Step3: Substitute into product - rule

$y'=u'v + uv'=\frac{1}{2}x^{-\frac{1}{2}}(x^{2}-1)+(x^{\frac{1}{2}} + 2)\cdot2x$.

Step4: Expand the expression

[ \begin{align*} y'&=\frac{1}{2}x^{-\frac{1}{2}}\cdot x^{2}-\frac{1}{2}x^{-\frac{1}{2}}+2x\cdot x^{\frac{1}{2}}+4x\ &=\frac{1}{2}x^{\frac{3}{2}}-\frac{1}{2}x^{-\frac{1}{2}}+2x^{\frac{3}{2}}+4x\ &=\left(\frac{1}{2}+2\right)x^{\frac{3}{2}}-\frac{1}{2}x^{-\frac{1}{2}}+4x\ &=\frac{5}{2}x^{\frac{3}{2}}-\frac{1}{2\sqrt{x}}+4x \end{align*} ]

Answer:

$\frac{5}{2}x^{\frac{3}{2}}-\frac{1}{2\sqrt{x}}+4x$