Verified answer

I am posting this wrong answer in the hope

that you or someone can see what I did wrong.

Suppose y = f(x) = ax^n with f(2) = 1 and f '(2) = 3

f(2) = a*2^n = 1

dy/dx = f'(x) = n*ax^(n - 1)

f'(2) = n*a2^(n - 1) = 3

dividing, (a*2^n)/[n*a2^(n - 1)] = 2/n = 1/3

n = 6

a*2^6 = 1

a = 1/64

y = x^6/64

dy/dx = 3x^5/32

dy/dx when x=3 is 729/32 ~ 22.78

Unfortunately this is wrong

It seemed OK until I looked at the part

f(x) = √(x+ f(³√(x^2 - 1)).

when x=3, the value of ³√(x^2 - 1) is 2, so this says

f(3) = √[3+ f(2)]

Based on y = x^6/64

f(2) would be 1 so

√[3+ f(2)] would be 2

whereas f(3) would be 3^6/64 which is clearly not 2.

Can anyone else put this right ?

Regards - Ian