Question

if u and v are differentiatable functions of x and if y = u.v then
$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$

Answer 1

y = u(x) v(x) YûL«PjRdL Nôo× Utßm
y′ = u(x) v′(x) + v(x) u′(x) … (14)
¨ìTQm: y = u(x) v(x)
y + ∆y = u(x + ∆x) v(x + ∆x)
∆y = u(x + ∆x) v(x + ∆x) − u(x) v(x)
∴ dy
dx = lim
∆x → 0
∆y
∆x

= lim
∆x → 0
u(x + ∆x) v(x + ∆x) − u(x) v(x)
∆x .
u(x + ∆x) v(x)-I ùRôϧ«p áh¥dL¯jÕ Uôt±VûUdL:
y′ = lim
∆x → 0
u(x + ∆x) v(x + ∆x) − u(x + ∆x) v(x) + u(x + ∆x) v(x) − u(x) v(x)
∆x
= lim
∆x → 0
u(x + ∆x) [ ] v(x + ∆x) − v(x) + v(x) [ ] u(x + ∆x) − u(x)
∆x
= lim
∆x → 0 u(x+∆x).
lim
∆x → 0 v(x + ∆x) − v(x)
∆x
+ v(x)
lim
∆x → 0
u(x + ∆x) − u(x)
∆x
ClúTôÕ u YûL«PjRdLRôp. CÕ ùRôPof£Vô]úR,
CYt±−ÚkÕ.
lim
∆x → 0 u(x + ∆x) = u(x)
u Utßm v B¡V] YûL«PjRdL Nôo×Ls GuTRôp
u′(x) =
lim
∆x → 0
u(x + ∆x) − u(x)
∆x
Utßm v′(x) =
lim
∆x → 0
v(x + ∆x) − v(x)
∆x .
G]úY y′ = u(x) v′(x) + v(x) u′(x).
AÕúTôX, u, v Utßm w GuT] YûL«PjRdL Nôo×Ls BL
CÚkÕ y = u(x) v(x) w(x) BLÜm CÚl©u
y′ = u(x) v(x) w′(x) + u(x) v′(x) w(x) + u′(x) v(x) w(x)
ϱl× (1) : úUtùNôu] ùTÚdLp ®§ûV ¸rdLôÔUôß ¨û]®p


This is product rule of differentiation