Question

How do you find the derivative of `f(t)=-2t^2+3t-6`?

Answer 1

`f(x)=-2t^2+3t-6 => f'(t)=-4t+3`

Explanation:

`f(x)=-2t^2+3t-6` is a polynomial

So, we must use the fact that the derivative of sums equals the sum of derivatives.

`f(x)=sum_(k=0)^na_kx^k => f'(x)=(sum_(k=0)^na_kx^k)'=sum_(k=0)^n(a_kx^k)'`

in this case

`f(x)=-2t^2+3t-6 => f'(x)=(-2t^2)'+(3t)'-(6)'`

Since 6 is a constant its derivative is zero. This is because the derivative of any constant is zero.

So,

`f'(x)=(-2t^2)'+(3t)'`

Next we can use the Power rule `(x^n)'=nx^(n-1)`
and the fact that a derivative times a constant equals constant times derivative `(cf(x))'=cf'(x)`

In this case `n=2` and `c=-2`

So

`(-2t^2)'=-2(t^2)'=-2(2)(t^(2-1))=-4t^1=-4t`

and

`(3t)'=3(t^1)'=3(t^(1-1))=3t^0=3(1)=3`

Then we put it together and get

`f'(x)=-4t+3`