1. ## Taking The Derivative

What is the derivative of this equation:

2. The chain rule is

$\frac{dy}{dt} = \frac{dy}{du}\times \frac{du}{dt}$

For your function $y = \arcsin(\sqrt{2t})$ make $u = \sqrt{2t}$ so $y = \arcsin(u)$

Now find $\frac{dy}{du}$ and $\frac{du}{dt}$

3. hi
just to add, $\forall x\in (-1,1),\arcsin'(x)=\frac{1}{\sqrt{1-x^2}}$

4. Then what is the derivative of (2t^.5)

5. When i took the derivative of (2t^.5) i got 1/(2t^.5)
does that even make sense?

6. Originally Posted by emilyroy10
Then what is the derivative of (2t^.5)
$\frac{d}{dt}(2 \sqrt{t}) = \frac{1}{\sqrt{t}}$

7. what is the derivative of ((2t)^.5)

8. $(\sqrt{2t})'=\frac{1}{2}(2t)^{-\frac{1}{2}}\times2=\frac{1}{\sqrt{2t}}$

9. okay thats what i got when i took the derivative of ((2t)^.5), which was 1/((2t)^.5).
but what do you get when you multiply it by the derivative of arcsin, which is 1/((1-((2t)^2))^.5)

10. you should get, $\frac{1}{\sqrt{2}\sqrt{1-2t}\sqrt{t}}$