# How do you find the derivative of the function f(x)=-2x^-3+x^2-7?

Aug 29, 2015

$\frac{\mathrm{df}}{\mathrm{dx}} = 6 {x}^{- 4} + 2 x$

#### Explanation:

The derivative of a sum of terms is equal to the sum of the derivatives of the individual terms:

$\frac{d \left(- 2 {x}^{3} + {x}^{2} - 7\right)}{\mathrm{dx}} = \frac{d \left(- 2 {x}^{-} 3\right)}{\mathrm{dx}} + \frac{d \left({x}^{2}\right)}{\mathrm{dx}} + \frac{d \left(- 7\right)}{\mathrm{dx}}$

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

So
$\textcolor{w h i t e}{\text{XXXX}} \frac{d \left(- 2 {x}^{- 3}\right)}{\mathrm{dx}} = \left(- 3\right) \cdot \left(- 2\right) {x}^{- 4} = 6 {x}^{- 4}$

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

$\textcolor{w h i t e}{\text{XXXX}} \frac{d \left(- 7\right)}{\mathrm{dx}} = \frac{d \left(- 7 {x}^{0}\right)}{\mathrm{dx}} = \left(0\right) \cdot \left(- 7\right) {x}^{- 1} = 0$