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

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Answer 1 · 2017-01-17T01:14:56

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

$f(x)=-2t^2+3t-6$ 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

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

How do you find the derivative of f(t)=-2t^2+3t-6f(t)=−2t2+3t−6f(t)=-2t^2+3t-6?