If f ( x ) + x 5 ( f ( x ) ) 2 = 34 and f ( 1 ) = 2 {"version":"1.1","math":"f(x)+x^5\left(f(x) ight)^2=34 ext{ and }f(1)=2"}
then f ′ ( 1 ) = {"version":"1.1","math":"f'(1)="}

Question

If  f ( x ) +  x 5    ( f ( x ) )  2  = 34  and  f ( 1 ) = 2 {"version":"1.1","math":"f(x)+x^5\left(f(x)ight)^2=34	ext{ and }f(1)=2"}
then   f ′  ( 1 ) = {"version":"1.1","math":"f'(1)="}

BlackTom AI · Accepted Answer

We start by restating the given information clearly and identifying the target derivative.
Given: f(x) + x^5 (f(x))^2 = 34, with f(1) = 2. We are asked to find f'(1).
Option analysis:
Option 1 (labeled 'Answer' in the provided choices): This option is incomplete as presented. The multiple-choice format expects a numeric value or a clearly defined expression, but the only option available is the word 'Answer', which does not convey a specific value. Therefore, it cannot be assessed as giving the correct numeric derivative.
To actually determine f'(1), differentiate the equation implicitly with respect to x:
- The derivative of f(x) is f'(x).
- The derivative of x^5 (f(x))^2 requires the product rule: d/dx[x^5] * (f(x))^2 + x^5 * d/dx[(f(x))^2]. This yields 5x^4 (f(x))^2 + x^5 * 2 f(x) f'(x).
Putting these together, we get: f'(x) + 5x^4 (f(x))^2 + x^5 * 2 f(x) f'(x) = 0.
Now evaluate at x = 1, using f(1) = 2:
- f'(1) + 5*(1)^4*(2)^2 + (1)^5 * 2*(2)*f'(1) = 0
- f'(1) + 5*4 + 4 f'(1) = 0
- (1)f'(1) + 4 f'(1) = -20
- 5 f'(1) = -20
- f'(1) = -4.
Thus, the actual numeric value for f'(1) would be -4.
However, the provided answer options do not present this numeric value; they only contain the placeholder 'Answer', which does not convey the correct derivative or any alternative option. This mismatch means the options do not reflect the correct conclusion, even though the calculation above shows the result that follows from the given equation and condition f(1) = 2.

If f ( x ) + x 5 ( f ( x ) ) 2 = 34 and f ( 1 ) = 2 {"version":"1.1","math":"f(x)+x^5\left(f(x)\right)^2=34\text{ and }f(1)=2"} then f ′ ( 1 ) = {"version":"1.1","math":"f'(1)="}

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