# Question #51bf7

Apr 13, 2016

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

#### Explanation:

Step 1 : we have $y = \frac{2 {x}^{2} - 1}{x \sqrt{1 + {x}^{2}}}$

This equation is of the form $f \left(x\right) = g \frac{x}{h \left(x\right)}$

Where, $g \left(x\right) = 2 {x}^{2} - 1$ and $h \left(x\right) = x \sqrt{1 + {x}^{2}}$

We will use the quotient rule to find the derivative.

$\implies f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - h ' \left(x\right) g \left(x\right)}{{h}^{2} \left(x\right)}$

Step 2 : Lets solve the different parts:

$g \left(x\right) = 2 {x}^{2} - 1$

$\implies \textcolor{red}{g ' \left(x\right) = 4 x}$

Step 3 : Next, to solve $h ' \left(x\right)$

We have $h \left(x\right) = x \sqrt{1 + {x}^{2}}$

This is of the form $h \left(x\right) = a \left(x\right) \cdot b \left(x\right)$

where, $a \left(x\right) = x$ and $b \left(x\right) = \sqrt{1 + {x}^{2}}$

Apply product rule:

$\implies \textcolor{b r o w n}{h ' \left(x\right) = a ' \left(x\right) b \left(x\right) + b ' \left(x\right) a \left(x\right)}$

$a \left(x\right) = x$

$\implies \textcolor{b l u e}{a ' \left(x\right) = 1}$

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

$\implies b \left(x\right) = {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}$

$\implies b ' \left(x\right) = \frac{1}{2} {\left(1 + {x}^{2}\right)}^{\frac{1}{2} - 1} \times 2 x$

Use $\textcolor{g r e e n}{f \left(x\right) = {x}^{n} \implies f ' \left(x\right) = n {x}^{n - 1}}$

$\implies b ' \left(x\right) = {\left(1 + {x}^{2}\right)}^{- \frac{1}{2}} \times x$

$\implies b ' \left(x\right) = \frac{1}{1 + {x}^{2}} ^ \left(\frac{1}{2}\right) \times x$

$\implies \textcolor{b l u e}{b ' \left(x\right) = \frac{1}{\sqrt{1 + {x}^{2}}} \times x}$

back to $\textcolor{b r o w n}{h ' \left(x\right) = a ' \left(x\right) b \left(x\right) + b ' \left(x\right) a \left(x\right)}$

Put back the values to find $h ' \left(x\right)$

$\implies h ' \left(x\right) = \sqrt{1 + {x}^{2}} + \frac{1}{\sqrt{1 + {x}^{2}}} \times x \times x$

$\implies h ' \left(x\right) = \frac{1 + {x}^{2} + {x}^{2}}{\sqrt{1 + {x}^{2}}}$

$\implies \textcolor{red}{h ' \left(x\right) = \frac{1 + 2 {x}^{2}}{\sqrt{1 + {x}^{2}}}}$

Final step : Put back the values in $f ' \left(x\right)$ and simplify:

$f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - h ' \left(x\right) g \left(x\right)}{{h}^{2} \left(x\right)}$

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

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

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

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

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

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

This looks long and confusing. But if you solve one step at a time, its a piece of cake.

All the best..