How to find dy/dt and dx/dt in ti84?

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I know how to find dy/dx from a point, but how can I find dy/dt and dx/dt from a point?

Please help me...my exam is in a few hours, and this is the last thing I need to learn
 
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Do you mean "find dy/dx at a point"? If so, you should be able to find dy/dt and dx/dt at a point if you have the formulas for x and y as functions of t.
 
Yes, I mean at a point. I wasn't given formulas...just points. I'll post an example when I get back home.
 
equation: y = sqrt(x)

find: a) dy/dt when x = 4
b) dx/dt when x = 25

given: dx/dt = 3

answers: a) 3/4
b) 20

I want to know how I can plug in the answers to make sure that they're correct. My exam is multiple choice and I don't have to show any work..
 
For a, using the chain rule, dy/dt = dy/dx * dx/dt
dy/dt|x = 4 = dy/dx|x = 4 * dx/dt|x = 4

You're given dx/dt = 3. Can you find dy/dx and then evaluate it when x = 4?

For b, I don't think you have given enough information.
 
Thanks for the help..now I will ace this exam for sure!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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