Asked by Anonymous
Let y=y(x)= x^8/f(x^2) and suppose f(1)=4 and f'(1)=5, find y'(-1).
Answers
Answered by
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.