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Use any method to evaluate the derivative of the following functions. $$h(r)=\frac{2-r-\sqrt{r}}{r+1}$$

Short Answer

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Question: Determine the derivative $$h'(r)$$ of the function $$h(r) = \frac{2 - r - \sqrt{r}}{r+1}$$. Answer: The derivative of the function is $$h'(r) = \frac{-r^{\frac{1}{2}}}{(r+1)^2}$$.

Step by step solution

01

Identify the given function

The given function is: $$h(r)=\frac{2-r-\sqrt{r}}{r+1}$$
02

Apply the Quotient Rule

Recall the quotient rule states: if we have a function $$\frac{f(r)}{g(r)},$$ its derivative is: $$\frac{f'(r)g(r)-f(r)g'(r)}{(g(r))^2}$$. So we need to differentiate both the numerator (f(r)) and the denominator (g(r)).
03

Differentiate the numerator (f(r))

In this function, $$f(r)=2-r-\sqrt{r}$$. We will differentiate each term with respect to r, starting with the constant, the linear term, and the square root term: $$\frac{d}{dr}(2) = 0$$ $$\frac{d}{dr}(-r) = -1$$ Now, we need to use the chain rule for the square root term (remembering that $$\sqrt{r}=r^{\frac{1}{2}}$$): $$\frac{d}{dr}(r^{\frac{1}{2}}) = \frac{1}{2}r^{-\frac{1}{2}}$$ So, $$f'(r)= 0 - 1 - \frac{1}{2}r^{-\frac{1}{2}}$$
04

Differentiate the denominator (g(r))

In this function, $$g(r) = r + 1$$. Differentiating with respect to r: $$\frac{d}{dr}(r+1) = 1$$ so, $$g'(r)=1$$
05

Apply the Quotient Rule

Now we have$$f'(r) = -1 - \frac{1}{2}r^{-\frac{1}{2}}$$and$$g'(r)=1,$$so we can find the derivative of h(r) using the quotient rule we identified earlier: $$h'(r)=\frac{f'(r)g(r)-f(r)g'(r)}{(g(r))^2}$$ $$=\frac{(-1-\frac{1}{2}r^{-\frac{1}{2}})(r+1)-(2-r-\sqrt{r})}{(r+1)^2}$$
06

Simplify the result

We can now distribute and simplify this expression: $$h'(r) = \frac{-1(r+1) - \frac{1}{2}r^{-\frac{1}{2}}(r+1) -2 + r +\sqrt{r}}{(r+1)^2}$$ $$= \frac{-r -1 -\frac{1}{2}r^{\frac{1}{2}}r -\frac{1}{2}r^{\frac{1}{2}} -2 +r+r^{\frac{1}{2}}}{(r+1)^2}$$ $$= \frac{-2r^{\frac{1}{2}}}{2(r+1)^2}$$ Thus, the derivative of the given function is $$h'(r) = \frac{-r^{\frac{1}{2}}}{(r+1)^2}$$

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