Question
Let y=y(x)= x^8/f(x^2) and suppose f(1)=4 and f'(1)=5, find y'(-1).
Answers
oobleck
since we have f(x^2), f(-1) = f(1)
y = x^8/f(x^2)
y' = 8x^7/f(x^2) - x^8/f^2(x^2) * f'(x^2) * 2x
so
y'(-1) = -8/4 + 5*2/16 = -11/8
y = x^8/f(x^2)
y' = 8x^7/f(x^2) - x^8/f^2(x^2) * f'(x^2) * 2x
so
y'(-1) = -8/4 + 5*2/16 = -11/8