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

Answer 1

f(x) + x²[f(x)]⁵ = 34

f(1) = 2

Differentiate using the product rule and the chain rule:

f'(x) + x² (5[f(x)]⁴) f'(x) + 2x [f(x)]⁵ = 0

Substitute x = 1:

f'(1) + (1)² (5[f(1)]⁴) f'(1) + 2(1) [f(1)]⁵ = 0

f'(1) + 5 [f(1)]⁴ f'(1) + 2 [f(1)]⁵ = 0

Substitute f(1) = 2:

f'(1) + 5 [2]⁴ f'(1) + 2 [2]⁵ = 0

f'(1) + 80 f'(1) + 64 = 0

81 f'(1) + 64 = 0

f'(1) = -64/81 = -0.79