Question

let ( f(x)=f(f(x)) ) and ( g(x)=(f(x))^{2} ).
you also know that ( f(6)=10, f(10)=2, f^{prime}(10)=4, f^{prime}(6)=8 )
find ( f^{prime}(6)= ) and ( g^{prime}(6)= )

Explanation:

Step1: Find \(F'(x)\) using the chain rule

The chain rule states that if \(y = f(u)\) and \(u = g(x)\), then \(y'=f'(u)\cdot g'(x)\). For \(F(x)=f(f(x))\), let \(u = f(x)\), so \(F'(x)=f'(f(x))\cdot f'(x)\)

Step2: Calculate \(F'(6)\)

Substitute \(x = 6\) into \(F'(x)\). We know that \(f(6)=10\) and \(f'(6) = 8\), \(f'(10)=4\). Then \(F'(6)=f'(f(6))\cdot f'(6)=f'(10)\cdot f'(6)\)

$$F'(6)=4\times8 = 32$$

Step3: Find \(G'(x)\) using the chain rule

Since \(G(x)=(F(x))^{2}\), by the chain rule \(G'(x)=2F(x)\cdot F'(x)\)

Step4: Calculate \(G'(6)\)

We already found \(F'(6) = 32\). Now substitute \(x = 6\) into \(G'(x)\). \(G'(6)=2F(6)\cdot F'(6)\). Since \(F(6)=f(f(6))=f(10) = 2\)

$$G'(6)=2\times2\times32=2048$$

Answer:

\(F'(6) = 32\) and \(G'(6) = 2048\)