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

Expert-verified Found in: Page 210 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Find the derivatives of the functions in Exercises 21–46. Keep in mind that it may be convenient to do some preliminary algebra before differentiating. $f\left(x\right)=x{\left(3{x}^{2}+1\right)}^{9}$

The required answer is ${\left(3{x}^{2}+1\right)}^{9}+54{x}^{2}{\left(3{x}^{2}+1\right)}^{8}$

See the step by step solution

## Step 1. Given Information

The given function is $f\left(x\right)=x{\left(3{x}^{2}+1\right)}^{9}$

## Step 2. Calculation

Differentiate both the sides with respect to x, we get,

$f\text{'}\left(x\right)=\frac{d}{dx}\left(x\right){\left(3{x}^{2}+1\right)}^{9}+x\left(\frac{d}{dx}{\left(3{x}^{2}+1\right)}^{9}\right)\phantom{\rule{0ex}{0ex}}={\left(3{x}^{2}+1\right)}^{9}+x\left(9{\left(3{x}^{2}+1\right)}^{9-1}\frac{d}{dx}\left(3{x}^{2}+1\right)\right)\phantom{\rule{0ex}{0ex}}={\left(3{x}^{2}+1\right)}^{9}+x\left(9{\left(3{x}^{2}+1\right)}^{8}\left(\frac{d}{dx}3{x}^{2}+\frac{d}{dx}1\right)\right)\phantom{\rule{0ex}{0ex}}={\left(3{x}^{2}+1\right)}^{9}+x\left(9{\left(3{x}^{2}+1\right)}^{8}\left(3\left(2x\right)\right)\right)\phantom{\rule{0ex}{0ex}}={\left(3{x}^{2}+1\right)}^{9}+54{x}^{2}{\left(3{x}^{2}+1\right)}^{8}$ ### Want to see more solutions like these? 