Question

need help with 1 step of a problem(calculus):
look at attached screenshot.
I am having trouble seeing how they get Dy'/dt = that big fraction.

Answer 1

your post left out the definition of x as a function of t, and y as a function of t

but do you see how to find dx/dt and dy/dt ?

Answer 2

Ah, I posted the actual problem now in this screenshot. I've gotten all the way to the part before the part that is highlighted in the first screenshot. I don't understand how they got the highlighted part

Answer 3

I think they are saying
\[ y' = \frac{dy}{dx} = \frac{-3 \sqrt{t+1} }{ \sqrt{3t}}\]
from the 2nd line.

take the derivative of y' with respect to t to get the next line
use the quotient rule
http://mathworld.wolfram.com/QuotientRule.html

Answer 4

so I would use dy/dt = 3/2(3t)^(-1/2) and take the derivative of that by using the quotient rule to get what is highlighted? Is there any way you can show step by step how to get the highlighted portion

Answer 5

you have
\[ y' = \frac{-3 \sqrt{t+1} }{ \sqrt{3t}} \]
find
\[ \frac{ d\ y' }{dt} \]
using
\[ d \frac{u}{v} = \frac{ v\ du - u\ dv }{v^2}\]
can you do that ?

Answer 6

\[ u = - 3(t+1)^{\frac{1}{2}} \]
and
\[ v= (3t)^{\frac{1}{2}} \]

Answer 7

*** so I would use dy/dt = 3/2(3t)^(-1/2) and take the derivative of that by using the quotient rule to get what is highlighted? ****

they are starting with dy/dx (not dy/dt) and taking the derivative with respect to t of dy/dx