Question

Derivadas:

$\frac{Delta F}{Delta T} = \frac{F(5+ Delta T)- F(5)}{Delta T} = \\ \\ \frac{Delta F}{Delta T} = \frac{[2000 - 10 (5+ Delta T)^2] - [2000-10*(5)^2]}{DeltaT} \\ \\ \frac{Delta F}{Delta T} = \frac{-100DeltaT - 10(DeltaT)^2}{DeltaT} = -100 - 10DeltaT$

De onde saiu -100 DeltaT?

Answer 1

delta T= d

lembrando o produto notavel: (a+b)² = a²+2ab+b²

$\frac{[2000-10(5+d)^2-[2000-10(5)^2]}{dt} \\\\ \frac{[2000-10(5^2+2*5*d+d^2)-[2000-10*25]}{d} \\\\ \frac{[2000-10(25+10d+d^2)-[2000-250]}{d} \\\\ \frac{[2000-250-100d-10d^2-1750]}{d} \\\\ \frac{-100d-10d^2}{d}= -100-10d$