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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?

2 Answers 2