If f(x) + x2[f(x)]5 = 34 and f(1) = 2, find f '(1).

Answer 1

f(x) + x²[f(x)]⁵ = 34

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

When x = 1, we have f'(1) + 2[f(1)]⁵ + 5[f(1)]⁴f'(1) = 0

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

81f'(1) = -64

f'(1) = -64/81