Taking The Derivative

**emilyroy10** · November 18th 2009, 12:15 PM

What is the derivative of this equation:
y= arcsin (radical(2t))

**pickslides** · November 18th 2009, 12:22 PM

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 $\frac{du}{dt}$

**Raoh** · November 18th 2009, 12:28 PM

hi

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

**emilyroy10** · November 18th 2009, 12:29 PM

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

**emilyroy10** · November 18th 2009, 12:32 PM

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

**e^(i*pi)** · November 18th 2009, 12:35 PM

Originally Posted by emilyroy10

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

$\frac{d}{dt}(2 \sqrt{t}) = \frac{1}{\sqrt{t}}$

**emilyroy10** · November 18th 2009, 12:36 PM

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

**Raoh** · November 18th 2009, 12:40 PM

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

**emilyroy10** · November 18th 2009, 12:43 PM

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)

**Raoh** · November 18th 2009, 12:49 PM

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

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