let ( f(x)=f(f(x)) ) and ( g(x)=(f(x))^{2} ).\nyou also know that ( f(6)=10, f(10)=2, f^{prime}(10)=4…

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

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

Answer

Answer:

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

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]