How do you differentiate (4x-3)/sqrt (2x^2+1)?

$\frac{4 x - 3}{\sqrt{2 {x}^{2} + 1}}$

First of all, we have a quotient. The derivative of a quotient $f \frac{x}{g} \left(x\right)$ is given by $\frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{{\left[g \left(x\right)\right]}^{2}}$. So we get that

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

Now we have to compute
$\frac{d}{\mathrm{dx}} \left[4 x - 3\right] = \frac{d}{\mathrm{dx}} \left[4 x\right] - \frac{d}{\mathrm{dx}} \left[3\right] = 4 \frac{d}{\mathrm{dx}} \left[x\right] + 0 = 4 \cdot 1 = 4$
and the derivative of a square root, which can be written as a power: $\sqrt{2 {x}^{2} + 1} = {\left(2 {x}^{2} + 1\right)}^{\frac{1}{2}}$. This is a composition of the square root with a second degree polynomial. The derivative of the composition $a \left(b \left(x\right)\right)$ is given by $a ' \left(b \left(x\right)\right) \cdot b ' \left(x\right)$. In this case $a \left(u\right) = {u}^{\frac{1}{2}}$ and $b \left(x\right) = 2 {x}^{2} + 1$, so
$\frac{d}{\mathrm{du}} \left[a \left(u\right)\right] = \frac{1}{2} {u}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{u}}$
$\frac{d}{\mathrm{dx}} \left[b \left(x\right)\right] = \frac{d}{\mathrm{dx}} \left[2 {x}^{2}\right] + \frac{d}{\mathrm{dx}} \left[1\right] = 2 \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] + 0 = 2 \cdot 2 x = 4 x$

So $\frac{d}{\mathrm{dx}} \left[a \left(b \left(x\right)\right)\right] = \frac{1}{2 \sqrt{2 {x}^{2} + 1}} \cdot 4 x = \frac{2 x}{\sqrt{2 {x}^{2} + 1}}$.

Finally we get

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