SOLUTION: [f(x)]^4 = (x + f(x))^3 and f(1)=2. What is f'(1)?

Question 1027271: [f(x)]^4 = (x + f(x))^3 and f(1)=2.
What is f'(1)?
Found 2 solutions by Fombitz, robertb:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!

I'm going to use y to simplify,

Implicitly differentiating,

However, it's not the case that because,

Actually,

has two real solutions,
and
which yields two values for the derivatives,
and
.
.
.
Here is the function along with the two tangent lines at
.
.
.
.

Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!

We will use implicit differentiation (a direct application of the chain rule) to find f'(1).
==> 4*f'(x) = 3*(1+f'(x)) by the chain rule.
==> 4*f'(1) = 3*(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.
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