# How do you calculate the derivative of (4x- 3)/(sqrt(2x^2 +1))?

Apr 2, 2015

You can use the product rule to find the derivative.

Rewrite the function as follows

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

Using the product rule

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(2 {x}^{2} + 1\right)}^{- \frac{3}{2}} \left[4 \left(2 {x}^{2} + 1\right) - 2 x \left(4 x - 3\right)\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(2 {x}^{2} + 1\right)}^{- \frac{3}{2}} \left[8 {x}^{2} + 4 - 8 {x}^{2} + 6 x\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(2 {x}^{2} + 1\right)}^{- \frac{3}{2}} \left[4 + 6 x\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {\left(2 {x}^{2} + 1\right)}^{- \frac{3}{2}} \left[2 + 3 x\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left(2 + 3 x\right)}{2 {x}^{2} + 1} ^ \left(\frac{3}{2}\right)$