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Q. 6

Expert-verifiedFound in: Page 209

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" $\frac{d}{dx}\left(f\right(u\left(x\right)\left)\right)$ without first finding the formula for $f\left(u\right(x\left)\right)$.

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

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:$f\left(u\right(x\left)\right)$

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

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 $x$:

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

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

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

Differentiate $\frac{df}{du}=\frac{(1-u)\frac{d}{du}({u}^{2}+3{u}^{5})-({u}^{2}+3{u}^{5})\frac{d}{du}(1-u)}{{(1-u)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{(1-u)(2u+15{u}^{4})-({u}^{2}+3{u}^{5})(-1)}{{(1-u)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2u+15{u}^{4}-2{u}^{2}-15{u}^{5}+{u}^{2}+3{u}^{5}}{{(1-u)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2u+15{u}^{4}-{u}^{2}-12{u}^{5}}{{(1-u)}^{2}}$

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

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

Substitute the values from step 3 and 4.

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

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

$\frac{d}{dx}\left(f\right(u\left(x\right)\left)\right)=\frac{2\sqrt{3{x}^{2}+1}+15{\left(\sqrt{3{x}^{2}+1}\right)}^{4}-{\left(\sqrt{3{x}^{2}+1}\right)}^{2}-12{\left(\sqrt{3{x}^{2}+1}\right)}^{5}}{{\left(1-\sqrt{3{x}^{2}+1}\right)}^{2}}\times \frac{3x}{\sqrt{3{x}^{2}+1}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3{x}^{2}+1}\left(2+15{\left(\sqrt{3{x}^{2}+1}\right)}^{3}-\sqrt{3{x}^{2}+1}-12{\left(\sqrt{3{x}^{2}+1}\right)}^{4}\right)}{{\left(1-\sqrt{3{x}^{2}+1}\right)}^{2}}\times \frac{3x}{\sqrt{3{x}^{2}+1}}\phantom{\rule{0ex}{0ex}}=\frac{3x\left(2+15{\left(\sqrt{3{x}^{2}+1}\right)}^{3}-\sqrt{3{x}^{2}+1}-12{\left(\sqrt{3{x}^{2}+1}\right)}^{4}\right)}{{\left(1-\sqrt{3{x}^{2}+1}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3x\left(2+15\left(3{x}^{2}+1\right)\sqrt{3{x}^{2}+1}-\sqrt{3{x}^{2}+1}-12{\left(3{x}^{2}+1\right)}^{2}\right)}{{\left(1-\sqrt{3{x}^{2}+1}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3x\left(2+45{x}^{2}\sqrt{3{x}^{2}+1}+15\sqrt{3{x}^{2}+1}-\sqrt{3{x}^{2}+1}-12\left(3{x}^{4}+1+6{x}^{2}\right)\right)}{{\left(1-\sqrt{3{x}^{2}+1}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3x\left(2+45{x}^{2}\sqrt{3{x}^{2}+1}+14\sqrt{3{x}^{2}+1}-36{x}^{4}-12-72{x}^{2}\right)}{{\left(1-\sqrt{3{x}^{2}+1}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3x\left(45{x}^{2}\sqrt{3{x}^{2}+1}+14\sqrt{3{x}^{2}+1}-36{x}^{4}-72{x}^{2}-10\right)}{{\left(1-\sqrt{3{x}^{2}+1}\right)}^{2}}\phantom{\rule{0ex}{0ex}}$

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