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Answers without the blur. Sign up and see all textbooks for free! Q. 23

Expert-verified Found in: Page 964 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Use Theorem 12.32 to find the indicated derivatives in Exercises 21–26. Express your answers as functions of a single variable. $\frac{dx}{dt}whenx=r\mathrm{cos}\theta ,r={t}^{2}-5,and\theta ={t}^{3}+1$

The value is $\frac{dx}{dt}=2t·\mathrm{cos}\left({t}^{3}+1\right)-3{t}^{2}\left({t}^{2}-5\right)\mathrm{sin}\left({t}^{3}+1\right)$

See the step by step solution

## Step 1. Given information:

Given:

$x=r\mathrm{cos}\theta ,\phantom{\rule{0ex}{0ex}}r={t}^{2}-5,and\phantom{\rule{0ex}{0ex}}\theta ={t}^{3}+1$

We have to find the indicated derivatives and express your answers as functions of a single variable.

## Step 2. Solution:

$U\mathrm{sin}gr={t}^{2}-5and\theta ={t}^{3}+1inx=r\mathrm{cos}\theta weget\phantom{\rule{0ex}{0ex}}x=\left({t}^{2}-5\right)·\mathrm{cos}\left({t}^{3}+1\right)\phantom{\rule{0ex}{0ex}}Diff.w.r.t.tweget\phantom{\rule{0ex}{0ex}}\frac{dx}{dt}=\left({t}^{2}-5\right)\frac{d}{dt}\mathrm{cos}\left({t}^{3}+1\right)+\mathrm{cos}\left({t}^{3}+1\right)\frac{d}{dt}\left({t}^{2}-5\right)\phantom{\rule{0ex}{0ex}}\frac{dx}{dt}=\left({t}^{2}-5\right)\left(-\mathrm{sin}\left({t}^{3}+1\right)·3{t}^{2}\right)+\mathrm{cos}\left({t}^{3}+1\right)·2t\phantom{\rule{0ex}{0ex}}\frac{dx}{dt}=2t·\mathrm{cos}\left({t}^{3}+1\right)-3{t}^{2}\left({t}^{2}-5\right)\mathrm{sin}\left({t}^{3}+1\right)$ ### Want to see more solutions like these? 