# How do you differentiate g(x)=6-5x^3?

Jun 7, 2017

$g ' \left(x\right) = - 15 {x}^{2}$

#### Explanation:

Differentiate each term:

$\frac{d}{\mathrm{dx}} \left[6\right] = 0$ (You should know that the derivative of any constant is always $0$)

Apply the Power rule: $\frac{d}{\mathrm{dx}} \left[{x}^{n}\right] = n {x}^{n - 1}$

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[- 5 {x}^{3}\right]$

$g ' \left(x\right) = - 5 \frac{d}{\mathrm{dx}} \left[{x}^{3}\right]$

$g ' \left(x\right) = - 5 \cdot 3 {x}^{3 - 1}$

$g ' \left(x\right) = - 5 \cdot 3 {x}^{2}$

$g ' \left(x\right) = - 15 {x}^{2}$

Jun 7, 2017

$f ' = - 15 {x}^{2}$

#### Explanation:

We use the power rule to differentiate the equation.
The power rule states $n {x}^{n - 1}$, where n is our exponent. We bring down the exponent and multiple it with our base $- 5 x$ in this case. So, $3 \left(- 5 x\right) = - 15 x$, then we subtract one from our original exponent so, ${x}^{3 - 1} = {x}^{2}$.
Regarding the 6, six is a constant the $\frac{d}{\mathrm{dx}}$ of a constant is always zero.

Our final answer is $f ' = 0 - 15 {x}^{2}$ or just $f ' = - 15 {x}^{2}$