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Problem 12 Medium Difficulty

Find the differential of each function.
(a) $ y = \frac {1 + 2u}{1 + 3u} $
(b) $ y = \theta^2 \sin 2\theta $

Answer

(a) $d y=\frac{-d u}{(1+3 u)^{2}}$
(b) $d y=2 \theta[\sin 2 \theta+\theta \cos (2 \theta)] d \theta$

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KX

Karl X.

October 15, 2019

Does this Video talking about this question?

Video Transcript

we're gonna be using the course people for this, which is a tough one. G minus F G one over G squared over geese. Words That's one plus three u Oh, it's just the denominator square. This simple 1st 2 negative one over one plus three u squared. Therefore d y his negative. Do you or negative one do you over one close three U square. Okay, Part B. We know we're gonna be using the product rule. So when I have to date us, sign tooth it up, plus tooth in a squared co sign tooth data on the reason why is because remember like the desk where is equivalent to theta Therefore, writing this in terms of do why we have to that we can write this infected form actually come signed to theta plus data pro sign to theatre