If frac{d}{d x}left{left(frac{x-1}{x-sqrt{x}} | vedclass

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(D) Let $y = \frac{x-1}{x-\sqrt{x}} e^{2x+1}$.We can simplify the expression $\frac{x-1}{x-\sqrt{x}}$ as follows:$\frac{x-1}{x-\sqrt{x}} = \frac{(\sqr

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$\frac{47}{24}$

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(D) Let $y = \frac{x-1}{x-\sqrt{x}} e^{2x+1}$.We can simplify the expression $\frac{x-1}{x-\sqrt{x}}$ as follows:$\frac{x-1}{x-\sqrt{x}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + \frac{1}{\sqrt{x}} = 1 + x^{-1/2}$.So, $y = (1 + x^{-1/2}) e^{2x+1}$.Now, differentiate $y$ with respect to $x$ using the product rule:$\frac{dy}{dx} = \frac{d}{dx}(1 + x^{-1/2}) \cdot e^{2x+1} + (1 + x^{-1/2}) \cdot \frac{d}{dx}(e^{2x+1})$.$\frac{dy}{dx} = (- \frac{1}{2} x^{-3/2}) e^{2x+1} + (1 + x^{-1/2}) \cdot 2 e^{2x+1}$.Factor out $e^{2x+1}$:$\frac{dy}{dx} = e^{2x+1} [-\frac{1}{2} x^{-3/2} + 2 + 2x^{-1/2}]$.We are given $\frac{dy}{dx} = \frac{x-1}{x-\sqrt{x}} e^{2x+1} f(x) = (1 + x^{-1/2}) e^{2x+1} f(x)$.Therefore, $f(x) = \frac{-\frac{1}{2} x^{-3/2} + 2 + 2x^{-1/2}}{1 + x^{-1/2}}$.Substitute $x = 4$:$f(4) = \frac{-\frac{1}{2} (4)^{-3/2} + 2 + 2(4)^{-1/2}}{1 + (4)^{-1/2}} = \frac{-\frac{1}{2} (\frac{1}{8}) + 2 + 2(\frac{1}{2})}{1 + \frac{1}{2}} = \frac{-\frac{1}{16} + 2 + 1}{\frac{3}{2}} = \frac{3 - \frac{1}{16}}{\frac{3}{2}} = \frac{\frac{47}{16}}{\frac{3}{2}} = \frac{47}{16} 	imes \frac{2}{3} = \frac{47}{24}$.

If $\frac{d}{d x}\left\{\left(\frac{x-1}{x-\sqrt{x}}\right) e^{2 x+1}\right\}=\frac{x-1}{x-\sqrt{x}} e^{2 x+1} f(x)$, then $f(4)=$

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