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

Jul 29, 2018

$f ' \left(x\right) = 6 {x}^{-} 4 + 2 x$

#### Explanation:

Use the power rule for differentiation:

We know that the derivative of ${x}^{n}$ is $n {x}^{n - 1}$.

The function $f \left(x\right) = - 2 {x}^{- 3} + {x}^{2} - 7$ has three parts.

We apply the power rule on each part.

color(red)(Part-1: $- 2 {x}^{-} 3$
Derivative: $- 2 \left(- 3\right) {x}^{\left(- 3 - 1\right)}$
Derivative: color(red)(6x^-4

color(blue)(Part-2: ${x}^{2}$
Derivative: $\left(2\right) {x}^{\left(2 - 1\right)}$
Derivative: color(blue)(2x

color(green)(Part-3: $7$
The derivative of a constant is zero, so:
Derivative: color(green)(0

We add the derivatives of each part up:

$\textcolor{red}{6 {x}^{-} 4} + \textcolor{b l u e}{2 x} + \textcolor{g r e e n}{0}$

So the derivative would be:

$f ' \left(x\right) = 6 {x}^{-} 4 + 2 x$

Jul 29, 2018

$6 {x}^{- 4} + 2 x$

#### Explanation:

Since we're dealing with a polynomial, we can find the derivative with the help of the Power Rule.

We simply multiply the exponent by the coefficient, and decrement the power by one. Recall that the derivative of a constant is zero.

We now have:

$6 {x}^{- 4} + 2 x$

Hope this helps!