Question

if y=sqrt(3x^2-3), determine dy/dt when x=4 and dx/dt=-1

Answer 1

i know that the derivative of \[\sqrt{x}=\frac{ 1 }{2\sqrt{x} }\]

Answer 2

@Jhannybean help? lol

Answer 3

you're off to a good start, but now you have to use the chain rule

Answer 4

what is the derivative of 3x^2-3 ?

Answer 5

6x

Answer 6

so using the chain rule gives you
\[\Large y=\sqrt{3x^2-3}\]
\[\Large \frac{dy}{dt}=\frac{1}{2\sqrt{3x^2-3}}*6x*\frac{dx}{dt}\]

Answer 7

first you derive sqrt(3x^2-3) as if you were deriving sqrt(x)

but then you have to derive the inside stuff 3x^2-3

and then you tack on dx/dt because when it comes to deriving the second inner stuff (the x) with respect to t, you get dx/dt

Answer 8

the last thing to do is to plug in x=4 and dx/dt=-1 and simplify

Answer 9

You are not missing the derivative wrt x on the left hand side, @jim_thompson5910 ?

Answer 10

you're thinking of

\[\Large \frac{dy}{dx} = \frac{dy/dt}{dx/dt}\]

\[\Large \frac{dy}{dx}*\frac{dx}{dt} = \frac{dy}{dt}\]

at least I think you are

Answer 11

Oh, okay.

Answer 12

but yeah that works too since the stuff in red is dy/dx

\[\Large \frac{dy}{dt} = {\color{red}{\frac{dy}{dx}}}*\frac{dx}{dt}\]

\[\Large \frac{dy}{dt} = {\color{red}{\frac{1}{2\sqrt{3x^2-3}}*6x}}*\frac{dx}{dt}\]

so either way is fine

Answer 13

the answer is \[\frac{ -4\sqrt{5} }{5}\]
Thanks :)

Answer 14

I'm getting the same thing. Nice job.