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

Question

Emma Aldridge · Accepted Answer

#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#

Ezekiel Alexander · Answer

undefined To find the derivative of $ f(t) = -2t^2 + 3t - 6 $, you would differentiate each term of the function separately using the power rule for derivatives. The power rule states that the derivative of $ x^n $ with respect to $ x $ is $ nx^{n-1} $.

Applying this rule to each term of $ f(t) $, we get:

$$ f'(t) = \frac{d}{dt}(-2t^2) + \frac{d}{dt}(3t) + \frac{d}{dt}(-6) $$

$$ f'(t) = -2 \cdot 2t^{2-1} + 3 \cdot 1t^{1-1} + 0 $$

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

So, the derivative of $ f(t) = -2t^2 + 3t - 6 $ with respect to $ t $ is $ f'(t) = -4t + 3 $.