Question

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

Answer 1

`g'(x)=-15x^2`

Explanation:

Differentiate each term:

`d/dx[6]=0` (You should know that the derivative of any constant is always `0`)

Apply the Power rule: `d/dx[x^n]=nx^(n-1)`

`g'(x)=d/dx[-5x^3]`

`g'(x)=-5d/dx[x^3]`

`g'(x)=-5*3x^(3-1)`

`g'(x)=-5*3x^2`

`g'(x)=-15x^2`

Answer 2

`f'=-15x^2`

Explanation:

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

Our final answer is `f'=0-15x^2` or just `f'=-15x^2`