How do you differentiate f(x)=(4x-3)/sqrt (2x^2+1)  using the quotient rule?

Dec 25, 2015

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

Explanation:

The quotient rule: if f(x)=g(x)/(h(x),

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

Thus,

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

Find each derivative:

$\textcolor{w h i t e}{s s s} \frac{d}{\mathrm{dx}} \left(4 x - 3\right) = 4$

Use the chain rule:

$\textcolor{w h i t e}{s s s} \frac{d}{\mathrm{dx}} \sqrt{2 {x}^{2} + 1} = \frac{1}{2} {\left(2 {x}^{2} + 1\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dx}} \left(2 {x}^{2} + 1\right)$

$\textcolor{w h i t e}{s s s} \implies \frac{1}{2 \sqrt{2 {x}^{2} + 1}} \cdot 4 x$

$\textcolor{w h i t e}{s s s} \implies \frac{2 x}{\sqrt{2 {x}^{2} + 1}}$

Plug these back into the equation for $f ' \left(x\right)$.

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

Multiply the numerator and denominator by $\sqrt{2 {x}^{2} + 1}$.

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

Simplify:

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