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Q. 28

Expert-verifiedFound in: Page 210

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.

$f\left(x\right)=\frac{{\left(1+\sqrt{x}\right)}^{2}}{3{x}^{2}-4x+1}$

The required answer is $\frac{1+\sqrt{x}}{\sqrt{x}(3{x}^{2}-4x+1)}-\frac{{\left(1+\sqrt{x}\right)}^{2}(6x-4)}{{\left(3{x}^{2}-4x+1\right)}^{2}}$

The given function is $f\left(x\right)=\frac{{\left(1+\sqrt{x}\right)}^{2}}{3{x}^{2}-4x+1}$

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

$f\text{'}\left(x\right)=\frac{\left(\frac{d}{dx}{\left(1+\sqrt{x}\right)}^{2}\right)\left(3{x}^{2}-4x+1\right)-{\left(1+\sqrt{x}\right)}^{2}\left(\frac{d}{dx}\left(3{x}^{2}-4x+1\right)\right)}{{\left(3{x}^{2}-4x+1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left(2\left(1+\sqrt{x}\right)\frac{d}{dx}\left(1+\sqrt{x}\right)\right)\left(3{x}^{2}-4x+1\right)-{\left(1+\sqrt{x}\right)}^{2}\left(3\left(2x\right)-4\right)}{{\left(3{x}^{2}-4x+1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left(2\left(1+\sqrt{x}\right)\frac{1}{2}{x}^{-\frac{1}{2}}\right)\left(3{x}^{2}-4x+1\right)-{\left(1+\sqrt{x}\right)}^{2}\left(6x)-4\right)}{{\left(3{x}^{2}-4x+1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{1+\sqrt{x}}{\sqrt{x}}\right)\left(3{x}^{2}-4x+1\right)-{\left(1+\sqrt{x}\right)}^{2}\left(6x)-4\right)}{{\left(3{x}^{2}-4x+1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left(1+\sqrt{x}\right)}{\sqrt{x}{\left(3{x}^{2}-4x+1\right)}^{2}}-\frac{{\left(1+\sqrt{x}\right)}^{2}\left(6x)-4\right)}{{\left(3{x}^{2}-4x+1\right)}^{2}}$

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