Question

let ( g ) be a function such that ( g(4)=8 ) and ( g^{prime}(4)=-3 ).
let ( h ) be the function ( h(x)=sqrt{x} ).
let ( h ) be a function defined as ( h(x)=g(x) cdot h(x) ).
( h^{prime}(4)= )

Explanation:

Step1: Find the derivative of \(H(x)\)

By the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = g(x)\) and \(v = h(x)=\sqrt{x}=x^{\frac{1}{2}}\).
The derivative of \(h(x)\) is \(h^\prime(x)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\).
So \(H^\prime(x)=g^\prime(x)\cdot h(x)+g(x)\cdot h^\prime(x)\).

Step2: Substitute \(x = 4\)

We know that \(g(4) = 8\), \(g^\prime(4)=-3\), and \(h(4)=\sqrt{4} = 2\), \(h^\prime(4)=\frac{1}{2\sqrt{4}}=\frac{1}{4}\).
Substitute these values into \(H^\prime(x)\):
\(H^\prime(4)=g^\prime(4)\cdot h(4)+g(4)\cdot h^\prime(4)\)
\(H^\prime(4)=(-3)\times2 + 8\times\frac{1}{4}\)

Step3: Calculate the value

First, calculate \((-3)\times2=-6\) and \(8\times\frac{1}{4} = 2\).
Then \(H^\prime(4)=-6 + 2\)

Answer:

\(-4\)