Question

Find the indicated derivative. In this case, the independent variable is a (unspecified) function of t.

Answer 1

y= x^1/2 + (1/x^1/2) , x=4 when t=1 , dx/dt t=1 = -8

Answer 2

your notation looks a bit confusing

Answer 3

sorry, let me try fixing it. \[y=\sqrt{x} + \frac{ 1 }{ \sqrt{x}} , x=4 when t=1, \frac{ dx }{ dt }| t=1| = -8\]

Answer 4

sorry, i don't understand this '-8' jazz at the end

Answer 5

It's okay, I don't either. I have no idea what any of this means!

Answer 6

anyone?

Answer 7

i suppose you need to find dy/dt ?

Answer 8

find dy/dx, then use
\(\large \dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}\)
where,
\(\dfrac{dt}{dx}= 1/(\dfrac{dx}{dt})=1/(-8)\)

Answer 9

I'm not quite sure how I'm supposed to use that

Answer 10

could you first find dy/dx ?

Answer 11

would it be : \[\frac{ .5 }{ x ^{.5} }- \frac{ .5 }{ x ^{1.5} }\] ?

Answer 12

yes, thats correct :)
now it would be easier if you first plug in x=4 before proceeding to: \(\large \dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}\)

Answer 13

\(\large \frac{ .5 }{ 4 ^{.5} }- \frac{ .5 }{ 4 ^{1.5} }=..?\)

Answer 14

okay found it, .1875? Not sure how to proceed after that

Answer 15

yes,
now at t=1, its given that dx/dt = -8
we found dy/dx = 0.1875
use \(\large \dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx} \implies 0.1875=\dfrac{dy}{dt}*(-8)\)
find dy/dt from here.

Answer 16

\[\frac{ -3 }{ 128 } \]

or am I completely lost? Cause according to my assignment that was incorrect

Answer 17

anyone know what I did wrong?