Derivative sirf calculate karna nahi — use karna bhi seekho! Rate of change, maxima, minima!
Ek factory owner hai jisko apna profit maximize karna hai. Uska profit function P(x) = -x² + 100x - 200 hai (x = units produced). Ab yeh function kab maximum value lega? Derivative lao: P'(x) = -2x + 100. P'(x) = 0 par x = 50. Aur P''(x) = -2 < 0, matlab yeh maximum point hai! So 50 units produce karna sabse zyada profitable hai. Yeh sab applications of derivatives hai!
dy/dx = rate of change of y with respect to x. Agar x time hai (t), toh dy/dt = rate of change of y over time.
Sphere ki radius 3 cm/sec ki rate se badh rahi hai. Volume ka rate of change find karo jab r = 5 cm.
V = (4/3)πr³ → dV/dt = 4πr² × dr/dt = 4π(25)(3) = 300π cm³/sec
Increasing: f'(x) > 0 on interval (a,b)
Decreasing: f'(x) < 0 on interval (a,b)
Critical points: f'(x) = 0 ya f'(x) undefined
f'(x) = 3x² - 3 = 3(x²-1) = 3(x-1)(x+1)
Critical points: x = -1, x = 1
x < -1: f'(x) > 0 → Increasing
-1 < x < 1: f'(x) < 0 → Decreasing
x > 1: f'(x) > 0 → Increasing
Local Max = f(1) = 1-6+9+15 = 19. Local Min = f(3) = 27-54+27+15 = 15
Q1. A stone is dropped into pond. Radius of ripple increases at 5 cm/s. Rate of increase of area when r=10 cm?
A=πr². dA/dt=2πr×dr/dt=2π(10)(5)=100π cm²/s
Q2. Find the tangent line to y=x³-3x at (2,2).
y'=3x²-3. At x=2: slope=12-3=9. Tangent: y-2=9(x-2) → y=9x-16
Q3. Among all rectangles of perimeter 100m, find dimensions for maximum area.
Let l+w=50 (half perimeter). A=lw=l(50-l). dA/dl=50-2l=0 → l=25. Square with side 25m → max area=625m²