# How do you find the derivative of 2x^2+x-1?

Jul 16, 2015

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

#### Explanation:

The derivative of $f \left(x\right) = p \left(x\right) + q \left(x\right) + r \left(x\right)$:
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{d \left(f \left(x\right)\right)}{\mathrm{dx}} = \frac{d \left(p \left(x\right)\right)}{\mathrm{dx}} + \frac{d \left(q \left(x\right)\right)}{\mathrm{dx}} + \frac{d \left(r \left(x\right)\right)}{\mathrm{dx}}$
That is, the derivative of a sum of (sub)functions is the sum of the derivatives of the (sub)functions.

The derivative of $a {x}^{b}$
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{d \left(a {x}^{b}\right)}{\mathrm{dx}} = b a {x}^{b - 1}$

Using
$\textcolor{w h i t e}{\text{XXXX}}$$p \left(x\right) = 2 {x}^{2}$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow \frac{d \left(p \left(x\right)\right)}{\mathrm{dx}} = 4 {x}^{1}$

$\textcolor{w h i t e}{\text{XXXX}}$$q \left(x\right) = x \left(= 1 {x}^{1}\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow \frac{d \left(q \left(x\right)\right)}{\mathrm{dx}} = \left(1\right) 1 {x}^{0} = 1$

$\textcolor{w h i t e}{\text{XXXX}}$$r \left(x\right) = - 1 \left(= \left(- 1\right) {x}^{0}\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow \frac{d \left(r \left(x\right)\right)}{\mathrm{dx}} = \left(0\right) \left(- 1\right) {x}^{- 1} = 0$

$\frac{d \left(f \left(x\right)\right)}{\mathrm{dx}} = 4 x + 1 \left(+ 0\right)$
$\textcolor{w h i t e}{\text{XXXX}}$when $f \left(x\right) = 2 {x}^{2} + x - 1$

(I recognize that this is a long explanation to a simple problem, but I wanted to try to cover any possible problems in following the solution).