If f(x) + x^(2)[f(x)]^5 = 34 and f(1) = 2, find f '(1).

Answer 1

f + x^2 f^5 = 34
f' + 2xf^5 + 5x^2 f^4 f' = 0
f' = -2xf^5/(1+5x^2 f^4)
f'(1) = -2(1)(2^5)/(1+5(1)(2^4))
= -64/80
= -4/5