# $\frac{d}{dt} \arcsin(\sqrt{2t})$

I'm trying to figure out how to solve this problem:

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

Wolfram Alpha gives me the following answer:

$$\frac{1}{\sqrt{2 - 4t}\sqrt{t}}$$

Here is what I've gotten:

$$y = \arcsin{\sqrt{2t}}$$ If $u = (2t)^{1/2}$ then $u' = \frac{1}{2} 2 t^{-1/2}$ therefore:

$$\frac{dy}{dt} = \frac{1}{\sqrt{1 - 2t}} \cdot \frac{1}{2 \sqrt{2t}}$$

Any Help?

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You didn't apply the chain rule when you took the derivative of $\sqrt{2t}$, in particular, there's a factor of 2 coming from differentiating $2t$.

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Differentiating the inverse trig' functions make use of the same trick. You have $y = \arcsin(\sqrt{2t})$, and so $\sin y = \sqrt{2t}$. We can use implicit differentiation on this: If $\sin y = \sqrt{2t}$ then

$$\cos y \frac{dy}{dt} = \frac{1}{\sqrt{2t}} \, .$$

Next we use the handy fact that $\sin^2 y + \cos^2y =1$ and couple that with that fact that $\sin y = \sqrt{2t}.$ We get $2t + \cos^2y = 1$ and in turn $\cos^2y = 1- 2t$. Assuming that $t\le 1/2$ gives $\cos y = \sqrt{1-2t}$. Finally, assuming that $0 < t < 1/2$ gives

$$\sqrt{1-2t}\frac{dy}{dt} = \frac{1}{\sqrt{2t}} \implies \frac{dy}{dt} = \frac{1}{\sqrt{2t(1-2t)}} \, .$$

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