University:
York University
Course:
MATH 1013 | Applied Calculus I
Academic year:

2022
Views:

86

Pages:

3
Author:

Nick T.

Derivatives and Rates of Change tangent: lim h→0 Δx Δy slope = tangent = Δx Δy Instantaneous Velocity Δy=Δs Δx=Δt lim Δt→0 Δt Δs Rate of change of a function at a point = limit of rate of change of the function over an interval Derivatives: study of change Limit of average rate of change f ′ (a)=lim h→0 h f(a+h)−f(a) f ′ (a)=lim x→a x−a f(x)−f(a) The definition of the derivative of a point Ex: f(x)= x lim h→0 h f(a+h)−f(a) lim h→0 h a+h − a $\lim_{h \to 0} \frac{\sqrt{a+h}-\sqrt{a}}{h} \cdot \frac{\sqrt{a+h}+\sqrt{a}}{\sqrt{a+h}+\sqrt{a}}$ lim h→0 h( a+h + a ) (a+h)−a lim h→0 a+h + a 1 2 a 1 1000 pairs sold per week Profit (production level) = 5000 + 100x Profit: 1,100? 10,500 1,500? 12,500 800? 9,000 h→0 lim h f(x+h)−f(x) h→0 lim h (x+h) 2 −3(x+h)−(x 2 −3x) = h→0 lim h x 2 +2xh+h 2 −3x−3h−x 2 +3x = h→0 lim h 2xh+h 2 −3h = h→0 lim h h(2x+h−3) =2x+0−3 =2x−3 f(x)= 2x+5 x−7 f ′ (x)= h→0 lim h f(x+h)−f(x) = h→0 lim h 2(x+h)+5 x+h−7 − 2x+5 x−7 = h→0 lim h(2x+2h+5)(2x+5) (x+h−7)(2x+5)−(x−7)(2x+2h+5) = h→0 lim h(2x+2h+5)(2x+5) (2x 2 +5x+2xh+5h−14x−35)−(2x 2 +2xh+5x−14x−35) = h→0 lim h(2x+2h+5)(2x+5) 19h = (2x+5) 2 19