hhhhhhhheeeeeelp please?

Question

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

how do i do this??? :(

Answer 1

f(x) + x^2[f(x)]^5 = 34

differentiating

f'(x) + x^2[5f(x)^4 f'(x)] + [f(x)]^5 (2x) = 0

f'(x) [1 + 5 x^2 f(x)^4] = - 2x [f(x)]^5

when x = 1

f'(1)[1 + 5f(1)^4] = -2[f(1)]^5

f'(1)[1 + 5(2)^4] = - 2[2]^5

f'(1) [ 81] = - 64

f'(1) = -64/81 =

Answer 2

f´(x) +2x*[f(x)]^5+x^2*5[f(x)]^4*f´(x)=0
Put in x=1
f´(1)+2*2^5+1*5*2^4*f´(1)=0
f´(1) = - 2^6/(1+5*2^4)