if ( f(x)=\frac{3 x^{2}+4 x + 7}{sqrt{x}} ), then:\n( f^{prime}(x)= )\n( f^{prime}(2)= )

Explanation:

Step1: Simplify the function

Rewrite (f(x)=\frac{3x^{2}+4x + 7}{\sqrt{x}}) as (f(x)=3x^{\frac{3}{2}}+4x^{\frac{1}{2}}+7x^{-\frac{1}{2}}) using the rule (\frac{a^{m}}{a^{n}}=a^{m - n}) ((a=x), (m = 2,1,0) and (n=\frac{1}{2})).

Step2: Differentiate term - by - term

Use the power rule ((x^{n})^\prime=nx^{n - 1}). For (y = 3x^{\frac{3}{2}}), (y^\prime=3\times\frac{3}{2}x^{\frac{3}{2}-1}=\frac{9}{2}x^{\frac{1}{2}}). For (y = 4x^{\frac{1}{2}}), (y^\prime=4\times\frac{1}{2}x^{\frac{1}{2}-1}=2x^{-\frac{1}{2}}). For (y = 7x^{-\frac{1}{2}}), (y^\prime=7\times(-\frac{1}{2})x^{-\frac{1}{2}-1}=-\frac{7}{2}x^{-\frac{3}{2}}). So (f^\prime(x)=\frac{9}{2}\sqrt{x}+\frac{2}{\sqrt{x}}-\frac{7}{2x\sqrt{x}}).

Step3: Evaluate (f^\prime(2))

Substitute (x = 2) into (f^\prime(x)). (f^\prime(2)=\frac{9}{2}\sqrt{2}+\frac{2}{\sqrt{2}}-\frac{7}{2\times2\times\sqrt{2}}). Rationalize the denominators: (\frac{2}{\sqrt{2}}=\sqrt{2}), (\frac{7}{4\sqrt{2}}=\frac{7\sqrt{2}}{8}). (f^\prime(2)=\frac{9\sqrt{2}}{2}+\sqrt{2}-\frac{7\sqrt{2}}{8}). Find a common denominator ((8)): (\frac{9\sqrt{2}\times4}{2\times4}+\frac{\sqrt{2}\times8}{1\times8}-\frac{7\sqrt{2}}{8}=\frac{36\sqrt{2}+16\sqrt{2}-7\sqrt{2}}{8}=\frac{45\sqrt{2}}{8}).

Answer:

(f^\prime(x)=\frac{9}{2}\sqrt{x}+\frac{2}{\sqrt{x}}-\frac{7}{2x\sqrt{x}}); (f^\prime(2)=\frac{45\sqrt{2}}{8})