Question 1027271
{{{(f(x))^4 = (x + f(x))^3}}}

We will use implicit differentiation (a direct application of the chain rule) to find f'(1).

==> 4{{{(f(x))^3}}}*f'(x) = 3{{{(x+f(x))^2}}}*(1+f'(x)) by the chain rule.

==> 4{{{(f(1))^3}}}*f'(1) = 3{{{(1+f(1))^2}}}*(1+f'(1)) after letting x = 1.

<==> 4*2^3*f'(1) = 3(1+2)^2*(1+f'(1))
<==>32f'(1) = 27(1+f'(1))
==> 32f'(1) = 27 + 27f'(1)
==> 5f'(1) = 27, or

f'(1) = 27/5.