Question

PLEASE HELP!

Answer 1

@surjithayer

Answer 2

@freckles

Answer 3

so far I have \(\Large\frac{-2x}{(1+x^2)^2}\), but I don't know what to do next.

Answer 4

\[\frac{dy}{dt}=\frac{dy}{dx} \cdot \frac{dx}{dt}\]

Answer 5

you just gave dy/dx

Answer 6

and you are given dx/dt

Answer 7

replace x with 2 now

Answer 8

ok, thanks

Answer 9

can oyu help with another? I'll tag you to it

Answer 10

did you want to tell me what you got when you evaluated dy/dt at x=2 first?

Answer 11

And I think I have time for one more question after this one.

Answer 12

yeah, I got -0.32

Answer 13

nice I got that too

Answer 14

:)

Answer 15

\[y=\frac{ 1 }{ 1+x^2 }\]
\[\frac{ dy }{ dt }=\frac{ 2x }{ \left( 1+x^2 \right)^2 }\frac{ dx }{ dt }\]
\[\frac{ dx }{ dt }=2 cm / \sec\]
when x=2
find dy/dt

Answer 16

similarly when x=-2
\[find~\frac{ dy }{ dt }\]