use any method to find the derivative of the function. h(r)=\frac{7 - 8r-sqrt{r}}{r + 1} h(r)=square

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $h(r)=\frac{u(r)}{v(r)}$, then $h^{\prime}(r)=\frac{u^{\prime}(r)v(r)-u(r)v^{\prime}(r)}{v(r)^2}$. Here, $u(r)=7 - 8r-\sqrt{r}=7 - 8r - r^{\frac{1}{2}}$ and $v(r)=r + 1$.

Step2: Find $u^{\prime}(r)$

Differentiate $u(r)$ term - by - term. The derivative of a constant $7$ is $0$, the derivative of $-8r$ is $-8$, and the derivative of $-r^{\frac{1}{2}}$ is $-\frac{1}{2}r^{-\frac{1}{2}}$. So, $u^{\prime}(r)=-8-\frac{1}{2\sqrt{r}}$.

Step3: Find $v^{\prime}(r)$

Differentiate $v(r)=r + 1$. The derivative of $r$ is $1$ and the derivative of the constant $1$ is $0$, so $v^{\prime}(r)=1$.

Step4: Apply the quotient - rule

[ \begin{align*} h^{\prime}(r)&=\frac{\left(-8-\frac{1}{2\sqrt{r}}\right)(r + 1)-(7 - 8r-\sqrt{r})\times1}{(r + 1)^2}\ &=\frac{-8r-8-\frac{r}{2\sqrt{r}}-\frac{1}{2\sqrt{r}}-7 + 8r+\sqrt{r}}{(r + 1)^2}\ &=\frac{-8r-8-\frac{\sqrt{r}}{2}-\frac{1}{2\sqrt{r}}-7 + 8r+\sqrt{r}}{(r + 1)^2}\ &=\frac{-15+\frac{\sqrt{r}}{2}-\frac{1}{2\sqrt{r}}}{(r + 1)^2}\ &=\frac{-15+\frac{r - 1}{2\sqrt{r}}}{(r + 1)^2}\ &=\frac{\frac{-30\sqrt{r}+r - 1}{2\sqrt{r}}}{(r + 1)^2}\ &=\frac{r-30\sqrt{r}-1}{2\sqrt{r}(r + 1)^2} \end{align*} ]

Answer:

$\frac{r-30\sqrt{r}-1}{2\sqrt{r}(r + 1)^2}$