Question

Find the derivative of g(t)=t^2 + t at t=-1 using the definition of the derivative at a point

Answer 1

Let me re-assemble this:
\[
g(t) = t^2 \\
g'(t) =\lim_{\Delta \to 0} \frac{g(t+\Delta) - g(t)}{\Delta} = \lim_{\Delta \to 0} \frac{(t+\Delta)^2 - (t)^2}{\Delta} =\\
= \lim_{\Delta \to 0} \frac{(t^2 +2\Delta t + \Delta^2) - t^2}{\Delta} = \lim_{\Delta \to 0} \frac{\Delta(2t + \Delta)}{\Delta} = \\
= \lim_{\Delta \to 0} 2t + \Delta = 2t \\
g'(-1) = 2 \cdot (-1) = -2
\]

Answer 2

Oh.. I missed +t sry... that's why I don't like people posting those as text =\
\[
g(t) = t^2 + t\\
g'(t) =\lim_{\Delta \to 0} \frac{g(t+\Delta) - g(t)}{\Delta} = \\
= \lim_{\Delta \to 0} \frac{\big[(t+\Delta)^2 + (t + \Delta) \big] - \big(t^2 + t\big)}{\Delta} =\\
= \lim_{\Delta \to 0} \frac{(t^2 +2\Delta t + \Delta^2 + t + \Delta) - t^2 - t}{\Delta} = \\
= \lim_{\Delta \to 0} \frac{2\Delta t + \Delta^2 + \Delta}{\Delta} = \lim_{\Delta \to 0} \frac{\Delta( 2t + \Delta + 1)}{\Delta} = \\\;\\
= \lim_{\Delta \to 0}\;2t + \Delta + 1 = 2t + 1\\
g'(-1) = 2 \cdot (-1) + 1 = -1
\]

Answer 3

Okay, got it! THanks