if ( f(x)=sqrt{4 x^{2}+7} ), find ( f^{prime}(5) ).

Explanation:

Step1: Rewrite the function

Rewrite ( f(x)=\sqrt{4x^{2}+7}=(4x^{2}+7)^{\frac{1}{2}})

Step2: Apply the chain - rule

The chain - rule states that if (y = u^{\frac{1}{2}}) and (u = 4x^{2}+7), then (y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}). First, find (\frac{dy}{du}): (\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}) Second, find (\frac{du}{dx}): (\frac{du}{dx}=8x) Then (f^\prime(x)=\frac{1}{2}(4x^{2}+7)^{-\frac{1}{2}}\cdot8x=\frac{4x}{\sqrt{4x^{2}+7}})

Step3: Substitute (x = 5)

Substitute (x = 5) into (f^\prime(x)): (f^\prime(5)=\frac{4\times5}{\sqrt{4\times5^{2}+7}}=\frac{20}{\sqrt{100 + 7}}=\frac{20}{\sqrt{107}})

Answer:

(\frac{20}{\sqrt{107}})