Question

if ( f(x)=\frac{7 x^{2}+7 x + 2}{sqrt{x}} ), then:
( f^{prime}(x)= )
( f^{prime}(5)= )
question help:

Explanation:

Step1: Rewrite the function

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

Step2: Apply the power rule

The power rule is \((x^{n})^\prime=nx^{n - 1}\).
For \(y = 7x^{\frac{3}{2}}\), \(y^\prime=7\times\frac{3}{2}x^{\frac{3}{2}-1}=\frac{21}{2}x^{\frac{1}{2}}\).
For \(y = 7x^{\frac{1}{2}}\), \(y^\prime=7\times\frac{1}{2}x^{\frac{1}{2}-1}=\frac{7}{2}x^{-\frac{1}{2}}\).
For \(y = 2x^{-\frac{1}{2}}\), \(y^\prime=2\times(-\frac{1}{2})x^{-\frac{1}{2}-1}=-x^{-\frac{3}{2}}\).
So \(f^\prime(x)=\frac{21}{2}x^{\frac{1}{2}}+\frac{7}{2}x^{-\frac{1}{2}}-x^{-\frac{3}{2}}=\frac{21x - x^{-1}+7}{2\sqrt{x}}\).

Step3: Calculate \(f^\prime(5)\)

Substitute \(x = 5\) into \(f^\prime(x)\).
\(f^\prime(5)=\frac{21\times5-\frac{1}{5}+7}{2\sqrt{5}}=\frac{\frac{525 - 1+35}{5}}{2\sqrt{5}}=\frac{\frac{559}{5}}{2\sqrt{5}}=\frac{559}{10\sqrt{5}}=\frac{559\sqrt{5}}{50}\approx\frac{559\times2.236}{50}\approx25.0\).

Answer:

\(f^\prime(x)=\frac{21x - x^{-1}+7}{2\sqrt{x}}\); \(f^\prime(5)=\frac{559\sqrt{5}}{50}\approx25.0\)