# How do you find the first and second derivatives of y=(2x^4-3x)/(4x-1) using the quotient rule?

Feb 29, 2016

First Derivative: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{24 {x}^{4} - 8 {x}^{3} + 3}{{\left(4 x - 1\right)}^{2}}$
Second Derivative: $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{192 {x}^{4} - 128 {x}^{3} + 24 {x}^{2} - 24}{4 x - 1} ^ 3$

#### Explanation:

First, I'll introduce the quotient rule.

If we are given $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$, then the derivative of $f \left(x\right)$ is:

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

Getting the first derivative:

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

Use the quotient rule.

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

The derivative of $2 {x}^{4} - 3 x$ is simply $8 {x}^{3} - 3$. The derivative of $4 x - 1$ is simply $4$. (I got these using power rule and difference rule)

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

Simplify.

$\left[4\right] \text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(32 {x}^{4} - 12 x - 8 {x}^{3} + 3\right) - \left(8 {x}^{4} - 12 x\right)}{{\left(4 x - 1\right)}^{2}}$

$\left[5\right] \text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{32 {x}^{4} - 12 x - 8 {x}^{3} + 3 - 8 {x}^{4} + 12 x}{{\left(4 x - 1\right)}^{2}}$

$\left[6\right] \text{ } \textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{24 {x}^{4} - 8 {x}^{3} + 3}{{\left(4 x - 1\right)}^{2}}}$

Getting the second derivative:

$\left[1\right] \text{ } \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left[\frac{24 {x}^{4} - 8 {x}^{3} + 3}{{\left(4 x - 1\right)}^{2}}\right]$

Use the quotient rule.

$\left[2\right] \text{ } \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{{\left(4 x - 1\right)}^{2} \cdot \frac{d}{\mathrm{dx}} \left(24 {x}^{4} - 8 {x}^{3} + 3\right) - \left(24 {x}^{4} - 8 {x}^{3} + 3\right) \cdot \frac{d}{\mathrm{dx}} \left[{\left(4 x - 1\right)}^{2}\right]}{{\left(4 x - 1\right)}^{2}} ^ 2$

The derivative of $24 {x}^{4} - 8 {x}^{3} + 3$ is $96 {x}^{3} - 24 {x}^{2}$. The derivative of ${\left(4 x - 1\right)}^{2}$ is $32 x - 8$.

$\left[3\right] \text{ } \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{{\left(4 x - 1\right)}^{2} \left(96 {x}^{3} - 24 {x}^{2}\right) - \left(24 {x}^{4} - 8 {x}^{3} + 3\right) \left(32 x - 8\right)}{4 x - 1} ^ 4$

Simplify.

$\left[4\right] \text{ } \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{{\left(4 x - 1\right)}^{2} \left(96 {x}^{3} - 24 {x}^{2}\right) - \left(24 {x}^{4} - 8 {x}^{3} + 3\right) \left(8\right) \left(4 x - 1\right)}{4 x - 1} ^ 4$

$\left[5\right] \text{ } \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{\cancel{\left(4 x - 1\right)} \left[\left(4 x - 1\right) \left(96 {x}^{3} - 24 {x}^{2}\right) - \left(24 {x}^{4} - 8 {x}^{3} + 3\right) \left(8\right)\right]}{4 x - 1} ^ \left(\cancel{4} 3\right)$

$\left[6\right] \text{ } \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{384 {x}^{4} - 96 {x}^{3} - 96 {x}^{3} + 24 {x}^{2} - 192 {x}^{4} + 64 {x}^{3} - 24}{4 x - 1} ^ 3$

$\left[7\right] \text{ } \textcolor{red}{\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{192 {x}^{4} - 128 {x}^{3} + 24 {x}^{2} - 24}{4 x - 1} ^ 3}$