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Numerade Educator

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Problem 13 Easy Difficulty

Find the differential of each function.
(a) $ y = \tan \sqrt t $
(b) $ y = \frac {1 - v^2}{1 + v^2} $

Answer

(a) $d y=\frac{\sec ^{2} \sqrt{t}}{2 \sqrt{t}} d t$
(b) $d y=\frac{-4 v}{\left(1+v^{2}\right)^{2}} d v$

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Video Transcript

using the channel. We are d y equals, seek it squared square root of tea over to squirt of tea. And then don't forget the DT on the end. Okay, use the quotient. Rules off one G minus. FT one over G squared. So just over the germ meter squared. This simplifies to negative for V over one plus v squared squared times dv.