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Answers without the blur. Sign up and see all textbooks for free! Q. 6

Expert-verified Found in: Page 209 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Suppose $u\left(x\right)=\sqrt{3{x}^{2}+1}$ and $f\left(u\right)=\frac{{u}^{2}+3{u}^{5}}{1-u}$. Use the chain rule to find role="math" localid="1648356625815" $\frac{d}{dx}\left(f\left(u\left(x\right)\right)\right)$ without first finding the formula for $f\left(u\left(x\right)\right)$.

The derivative of the function is:$\frac{3x\left(45{x}^{2}\sqrt{3{x}^{2}+1}+14\sqrt{3{x}^{2}+1}-36{x}^{4}-72{x}^{2}-10\right)}{{\left(1-\sqrt{3{x}^{2}+1}\right)}^{2}}$

See the step by step solution

## Step 1. Given information:

The functions are:

$u\left(x\right)=\sqrt{3{x}^{2}+1}$

$f=\frac{{u}^{2}+3{u}^{5}}{1-u}$

The composite function is:$f\left(u\left(x\right)\right)$

## Step 2. Find derivative of u(x) using chain rule.

Since, $u\left(x\right)=\sqrt{3{x}^{2}+1}\phantom{\rule{0ex}{0ex}}u\left(x\right)={\left(3{x}^{2}+1\right)}^{\frac{1}{2}}$

Let $v=3{x}^{2}+1$

Then $u={\left(v\right)}^{\frac{1}{2}}$

Derivative of $u\left(v\right)$ with respect to $v$:

$\frac{du}{dv}=\frac{1}{2}{\left(v\right)}^{\frac{1}{2}-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{\left(v\right)}^{-\frac{1}{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2\sqrt{v}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2\sqrt{3{x}^{2}+1}}$

Derivative of $v\left(x\right)$ with respect to $x$:

$\frac{dv}{dx}=\frac{d}{dx}\left(3{x}^{2}+1\right)\phantom{\rule{0ex}{0ex}}=3.2x+0\phantom{\rule{0ex}{0ex}}=6x$

$\frac{du}{dx}=\frac{du}{dv}×\frac{dv}{dx}\phantom{\rule{0ex}{0ex}}=\frac{1}{2\sqrt{3{x}^{2}+1}}×6x\phantom{\rule{0ex}{0ex}}=\frac{3x}{\sqrt{3{x}^{2}+1}}$

## Step 3. Find the derivative of f(u):

$f\left(u\right)=\frac{{u}^{2}+3{u}^{5}}{1-u}$

Differentiate

## Step 4. Find derivative of f(u(x)):

By using the chain rule, the derivative of the function is:

$\frac{d}{dx}\left(f\left(u\left(x\right)\right)\right)=\frac{df}{du}×\frac{du}{dx}$

Substitute the values from step 3 and 4.

$\frac{d}{dx}\left(f\left(u\left(x\right)\right)\right)=\frac{2u+15{u}^{4}-{u}^{2}-12{u}^{5}}{{\left(1-u\right)}^{2}}×\frac{3x}{\sqrt{3{x}^{2}+1}}$

Substitute value of $u\left(x\right)=\sqrt{3{x}^{2}+1}$ and simplify:

$\frac{d}{dx}\left(f\left(u\left(x\right)\right)\right)=\frac{2\sqrt{3{x}^{2}+1}+15{\left(\sqrt{3{x}^{2}+1}\right)}^{4}-{\left(\sqrt{3{x}^{2}+1}\right)}^{2}-12{\left(\sqrt{3{x}^{2}+1}\right)}^{5}}{{\left(1-\sqrt{3{x}^{2}+1}\right)}^{2}}×\frac{3x}{\sqrt{3{x}^{2}+1}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3{x}^{2}+1}\left(2+15{\left(\sqrt{3{x}^{2}+1}\right)}^{3}-\sqrt{3{x}^{2}+1}-12{\left(\sqrt{3{x}^{2}+1}\right)}^{4}\right)}{{\left(1-\sqrt{3{x}^{2}+1}\right)}^{2}}×\frac{3x}{\sqrt{3{x}^{2}+1}}\phantom{\rule{0ex}{0ex}}=\frac{3x\left(2+15{\left(\sqrt{3{x}^{2}+1}\right)}^{3}-\sqrt{3{x}^{2}+1}-12{\left(\sqrt{3{x}^{2}+1}\right)}^{4}\right)}{{\left(1-\sqrt{3{x}^{2}+1}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3x\left(2+15\left(3{x}^{2}+1\right)\sqrt{3{x}^{2}+1}-\sqrt{3{x}^{2}+1}-12{\left(3{x}^{2}+1\right)}^{2}\right)}{{\left(1-\sqrt{3{x}^{2}+1}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3x\left(2+45{x}^{2}\sqrt{3{x}^{2}+1}+15\sqrt{3{x}^{2}+1}-\sqrt{3{x}^{2}+1}-12\left(3{x}^{4}+1+6{x}^{2}\right)\right)}{{\left(1-\sqrt{3{x}^{2}+1}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3x\left(2+45{x}^{2}\sqrt{3{x}^{2}+1}+14\sqrt{3{x}^{2}+1}-36{x}^{4}-12-72{x}^{2}\right)}{{\left(1-\sqrt{3{x}^{2}+1}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3x\left(45{x}^{2}\sqrt{3{x}^{2}+1}+14\sqrt{3{x}^{2}+1}-36{x}^{4}-72{x}^{2}-10\right)}{{\left(1-\sqrt{3{x}^{2}+1}\right)}^{2}}\phantom{\rule{0ex}{0ex}}$ ### Want to see more solutions like these? 