Question

find y' dy where y = 2/x^3+x
--
dx x=1

Answer 1

y = 2/x^3+x
dy/dx = 2/3x^2+1
x = 1
(2/1^3)+1
= 3

Answer 2

\[y=\frac{2}{x^3}+x?\]

Answer 3

yes and dy/dx x=1

Answer 4

\[y=2x^{-3}+x => y'=2(-3)x^{-3-1}+1=-6x^{-4}+1=\frac{-6}{x^4}+1\]

Answer 5

\[y'|_{x=1}=\frac{-6}{(1)^4}+1=-6+1=-5\]

Answer 6

awesome I also got -5 but was not sure if I did it right

Answer 7

great!

Answer 8

thx so much

Answer 9

np :)