test ii. use delta (δ) - method\n3. find the first derivative with respect to x, $y = \\frac{2x +…

Explanation:

Step1: Recall quotient - rule for 3

The quotient rule for $y=\frac{u}{v}$ is $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. For $y = \frac{2x + 3}{\sqrt{x+2}}$, let $u = 2x+3$ and $v=(x + 2)^{\frac{1}{2}}$. $u^\prime=2$ and $v^\prime=\frac{1}{2}(x + 2)^{-\frac{1}{2}}$.

Step2: Apply quotient - rule for 3

[ \begin{align*} y^\prime&=\frac{2\sqrt{x + 2}-(2x + 3)\frac{1}{2}(x + 2)^{-\frac{1}{2}}}{x+2}\ &=\frac{2(x + 2)-(2x + 3)\frac{1}{2}}{(x + 2)^{\frac{3}{2}}}\ &=\frac{4(x + 2)-(2x + 3)}{2(x + 2)^{\frac{3}{2}}}\ &=\frac{4x+8 - 2x-3}{2(x + 2)^{\frac{3}{2}}}\ &=\frac{2x + 5}{2(x + 2)^{\frac{3}{2}}} \end{align*} ]

Step3: Rewrite the function for 4

For $xy=\sqrt{x^{2}-x - 2}$, we can write $y=\frac{\sqrt{x^{2}-x - 2}}{x}= \frac{(x^{2}-x - 2)^{\frac{1}{2}}}{x}$. Let $u=(x^{2}-x - 2)^{\frac{1}{2}}$ and $v = x$. Then $u^\prime=\frac{2x - 1}{2(x^{2}-x - 2)^{\frac{1}{2}}}$ and $v^\prime=1$.

Step4: Apply quotient - rule for 4

[ \begin{align*} y^\prime&=\frac{\frac{2x - 1}{2(x^{2}-x - 2)^{\frac{1}{2}}}\cdot x-(x^{2}-x - 2)^{\frac{1}{2}}\cdot1}{x^{2}}\ &=\frac{\frac{x(2x - 1)-2(x^{2}-x - 2)}{2(x^{2}-x - 2)^{\frac{1}{2}}}}{x^{2}}\ &=\frac{2x^{2}-x-2x^{2}+2x + 4}{2x^{2}(x^{2}-x - 2)^{\frac{1}{2}}}\ &=\frac{x + 4}{2x^{2}(x^{2}-x - 2)^{\frac{1}{2}}} \end{align*} ]

Step5: Apply quotient - rule again for 4 (to find second - derivative)

Let $u=x + 4$, $u^\prime=1$, $v = 2x^{2}(x^{2}-x - 2)^{\frac{1}{2}}$. First, find $v^\prime$ using the product rule $(ab)^\prime=a^\prime b+ab^\prime$ where $a = 2x^{2}$ and $b=(x^{2}-x - 2)^{\frac{1}{2}}$. $a^\prime=4x$, $b^\prime=\frac{2x - 1}{2(x^{2}-x - 2)^{\frac{1}{2}}}$. [ \begin{align*} v^\prime&=4x(x^{2}-x - 2)^{\frac{1}{2}}+2x^{2}\cdot\frac{2x - 1}{2(x^{2}-x - 2)^{\frac{1}{2}}}\ &=\frac{4x(x^{2}-x - 2)+2x^{2}(2x - 1)}{(x^{2}-x - 2)^{\frac{1}{2}}}\ &=\frac{4x^{3}-4x^{2}-4x + 4x^{3}-2x^{2}}{(x^{2}-x - 2)^{\frac{1}{2}}}\ &=\frac{8x^{3}-6x^{2}-4x}{(x^{2}-x - 2)^{\frac{1}{2}}} \end{align*} ] Then, by the quotient rule $y^{\prime\prime}=\frac{u^\prime v-uv^\prime}{v^{2}}$. [ \begin{align*} y^{\prime\prime}&=\frac{2x^{2}(x^{2}-x - 2)^{\frac{1}{2}}-(x + 4)\frac{8x^{3}-6x^{2}-4x}{(x^{2}-x - 2)^{\frac{1}{2}}}}{4x^{4}(x^{2}-x - 2)}\ &=\frac{2x^{2}(x^{2}-x - 2)-(x + 4)(8x^{3}-6x^{2}-4x)}{4x^{4}(x^{2}-x - 2)^{\frac{3}{2}}} \end{align*} ]

Answer:

For question 3: $y^\prime=\frac{2x + 5}{2(x + 2)^{\frac{3}{2}}}$ For question 4: $y^{\prime\prime}=\frac{2x^{2}(x^{2}-x - 2)-(x + 4)(8x^{3}-6x^{2}-4x)}{4x^{4}(x^{2}-x - 2)^{\frac{3}{2}}}$