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

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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) About Me  (Show Source):
You can put this solution on YOUR website!

I'm going to use y to simplify,
y%5E4=x%5E3%2B3x%5E2y%2B3xy%5E2%2By%5E3
Implicitly differentiating,
4y%5E3dy=3x%5E2dx%2B3%28x%5E2dy%2B2xydx%29%2B3%28y%5E2dx%2B2xydy%29%2B3y%5E2dy
4y%5E3dy-3y%5E2dy-3x%5E2dy-6xydy%29=3x%5E2dx%2B6xydx%2B3y%5E2dx
%284y%5E3-3y%5E2-6xy-3y%5E2%29dy=3%28x%5E2%2B2xy%2By%5E2%29dx
%284y%5E3-3y%5E2-6xy-3y%5E2%29dy=3%28x%2By%29%5E2dx
dy%2Fdx=%283%28x%2By%29%5E2%29%2F%284y%5E3-3y%5E2-6xy-3y%5E2%29
However, it's not the case that f%281%29=2 because,
%282%29%5E4=%281%2B2%29%5E3
16=%283%29%5E3
16%3C%3E27
Actually,
y%5E4=%281%2By%29%5E3
has two real solutions,
y=-0.55 and y=2.63
which yields two values for the derivatives,
dy%2Fdx=-0.480 and dy%2Fdx=1.189
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Here is the function along with the two tangent lines at x=1
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Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
%28f%28x%29%29%5E4+=+%28x+%2B+f%28x%29%29%5E3
We will use implicit differentiation (a direct application of the chain rule) to find f'(1).
==> 4%28f%28x%29%29%5E3*f'(x) = 3%28x%2Bf%28x%29%29%5E2*(1+f'(x)) by the chain rule.
==> 4%28f%281%29%29%5E3*f'(1) = 3%281%2Bf%281%29%29%5E2*(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.