An Alternative Proof of the chain rule and definition of differentiability for multivariable functions
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This document presents an alternative proof of the chain rule for multivariable functions. It begins by defining the chain rule formula and then proves it using a new definition of differentiability. The proof shows that under the new definition, differentiability implies continuity and satisfies the chain rule. It also proves that the old definition of differentiability implies the new one, so both definitions are equivalent. The conclusion is that the old definition of differentiability is sufficient to prove the chain rule.
![Definition: f is [new] differentiable at r0 if
f ( x, y ) − f ( x0 , y ) ∂f
lim
=
( r0 )
r →r0
x − x0
∂x
and
f ( x, y ) − f ( x, y 0 ) ∂f
lim
=
( r0 )
r →r0
y − y0
∂y](https://image.slidesharecdn.com/chainrule-131113091337-phpapp01/85/An-Alternative-Proof-of-the-chain-rule-and-definition-of-differentiability-for-multivariable-functions-7-320.jpg)
![Definition: f is [old] differentiable at r0 if
∆f ( r0 ) = f ( r0 + ∆r ) − f ( r0 )
∂f
∂f
=
( r0 ) ∆x + ( r0 ) ∆y + ε1∆x + ε 2∆y
∂x
∂y
where ε1 = ε 1 ( ∆r ) , ε 2 = ε 2 ( ∆r ) → 0 as ∆r → 0
∆x = x − x0 , ∆y = y − y 0 , ∆r = r − r0](https://image.slidesharecdn.com/chainrule-131113091337-phpapp01/85/An-Alternative-Proof-of-the-chain-rule-and-definition-of-differentiability-for-multivariable-functions-8-320.jpg)
An Alternative Proof of the chain rule and definition of differentiability for multivariable functions
- 1. An alternative proof for the chain rule for multivariable functions Raymond Jensen Northern State University
- 2. Chain rule: let f be differentiable wrt. x, y, and x, y differentiable wrt. t. x = x(t) y = y (t) f = f ( x, y ) df ∂f dx ∂f dy = + dt ∂x dt ∂y dt
- 3. Proof g ( t ) − g ( t0 ) dg ( t0 ) = lim t →t0 dt t − t0 f ( x ( t ) ,y ( t ) ) = g ( t ) =g ( t ) df ( x ( t0 ) , y ( t0 ) ) = lim0 t →t dt =g ( t0 ) 64 744 64 744 4 8 4 8 f ( x ( t ) , y ( t ) ) − f ( x ( t0 ) , y ( t0 ) ) t − t0
- 4. Proof: begin as with the product rule df df x ( t0 ) , y ( t0 ) ) = ( r0 ) ( dt dt f ( x ( t ) , y ( t ) ) − f ( x ( t0 ) , y ( t ) ) + f ( x ( t0 ) , y ( t ) ) − f ( x ( t0 ) , y ( t0 ) ) = lim t →t0 t − t0 = lim t →t0 f ( x ( t ) , y ( t ) ) − f ( x ( t0 ) , y ( t ) ) t − t0 + lim t →t0 f ( x ( t0 ) , y ( t ) ) − f ( x ( t0 ) , y ( t 0 ) ) t − t0
- 5. Proof: continue as with the single-variable chain rule f ( x ( t ) , y ( t ) ) − f ( x ( t0 ) , y ( t ) ) x ( t ) − x ( t0 ) df ( r0 ) = lim t →t0 dt x ( t ) − x ( t0 ) t − t0 f ( x ( t0 ) , y ( t ) ) − f ( x ( t0 ) , y ( t0 ) ) y ( t ) − y ( t0 ) + lim t →t0 y ( t ) − y ( t0 ) t − t0 f ( x, y ) − f ( x0 , y ) x ( t ) − x ( t0 ) = lim lim r →r0 t →t0 x − x0 t − t0 f ( x0 , y ) − f ( x 0 , y 0 ) y ( t ) − y ( t0 ) + lim lim r →r0 t →t0 y − y0 t − t0
- 6. Proof: continue as with the single-variable chain rule f ( x, y ) − f ( x 0 , y ) x ( t ) − x ( t0 ) df lim ( r0 ) = rlim →r0 t →t0 dt x − x0 t − t0 144424443 144 244 3 =" ∂f " ∂x = dx dt f ( x0 , y ) − f ( x0 , y 0 ) y ( t ) − y ( t0 ) + lim lim r →r0 t →t0 y − y0 t − t0 1444 24444 144244 4 3 3 = ∂f ∂y = dy dt
- 7. Definition: f is [new] differentiable at r0 if f ( x, y ) − f ( x0 , y ) ∂f lim = ( r0 ) r →r0 x − x0 ∂x and f ( x, y ) − f ( x, y 0 ) ∂f lim = ( r0 ) r →r0 y − y0 ∂y
- 8. Definition: f is [old] differentiable at r0 if ∆f ( r0 ) = f ( r0 + ∆r ) − f ( r0 ) ∂f ∂f = ( r0 ) ∆x + ( r0 ) ∆y + ε1∆x + ε 2∆y ∂x ∂y where ε1 = ε 1 ( ∆r ) , ε 2 = ε 2 ( ∆r ) → 0 as ∆r → 0 ∆x = x − x0 , ∆y = y − y 0 , ∆r = r − r0
- 9. Old differentiability ⇒ chain rule ∆f ∂f ∆x ∂f ∆y ∆x ∆y ( r0 ) = ( r0 ) + ( r0 ) + ε1 + ε 2 ∆t ∂x ∆t ∂y ∆t ∆t ∆t df ∂f dx ∂f dy → ( r0 ) = ( r0 ) + ( r0 ) dt ∂x dt ∂y dt
- 10. Old differentiability ⇒ continuity • Leithold thm. 12.4.3
- 11. New differentiability ⇒ continuity lim f ( r ) − f ( r0 ) = lim f ( x, y ) − f ( x, y 0 ) + f ( x, y 0 ) − f ( x0 , y 0 ) r →r0 r →r0 = lim f ( x, y ) − f ( x, y 0 ) + lim f ( x, y 0 ) − f ( x0 , y 0 ) r →r0 r →r0 f ( x, y ) − f ( x , y 0 ) = lim lim r →r0 ( y − y 0 ) r →r0 y − y0 1444 24444 14243 4 3 =0 = ∂f ∂y f ( x, y 0 ) − f ( x0 , y 0 ) + lim lim r →r0 ( x − x0 ) r →r0 x − x0 14444 24444 14243 3 =0 = =0 ∂f ∂x
- 12. If partials of f exist near r0 and are continuous at r0 then f is old differentiable at r0. • Stewart thm. 11.4.8, Leithold thm. 12.4.4
- 13. If partials of f exist near r0 and are continuous at r0 then f is new differentiable at r0. f ( x, y ) − f ( x0 , y ) ∂f f ( x, y ) − f ( x 0 , y ) f ( x, y 0 ) − f ( x 0 , y 0 ) lim − − lim ( r0 ) = rlim r →r0 →r0 r →r0 x − x0 ∂x x − x0 x − x0 ( ) 678 6 74 6 74 6 74 4( 0 )8 4 ( ) 8 4( 0 ) 8 f ( x, y ) − f ( x 0 , y ) − f ( x , y 0 ) + f ( x 0 , y 0 ) = lim r →r0 x − x0 g x g x h x g ( x ) − g ( x0 ) − h ( x ) − h ( x 0 ) = lim r →r0 x − x0 MVT h x
- 14. If partials exist near r0 and are continuous at r0 then f is new differentiable at r0. f ( x, y ) − f ( x0 , y ) ∂f g ′ ( u ) ∆x − h′ ( v ) ∆x lim − ( r0 ) = rlim r →r0 →r0 x − x0 ∂x ∆x = lim g ′ ( u ) − h′ ( v ) r →r0 ∂f ∂f = lim ( u, y ) − lim ( v , y 0 ) r →r0 ∂x r →r0 ∂x ∂f ∂f = ( r0 ) − ( r0 ) ∂x ∂x =0
- 15. Old differentiability ⇒ new differentiability f ( x, y ) − f ( x, y 0 ) ∂f lim − ( r0 ) r →r0 y − y0 ∂y f ( x, y ) − f ( x, y 0 ) − f ( x0 , y ) + f ( x0 , y 0 ) = lim r →r0 y − y0 f ( x, y ) − f ( x0 , y 0 ) − f ( x, y 0 ) − f ( x0 , y ) + 2f ( x0 , y 0 ) = lim r →r0 y − y0 f ( x, y ) − f ( x 0 , y 0 ) f ( x 0 , y 0 ) − f ( x, y 0 ) f ( x0 , y 0 ) − f ( x0 , y ) = lim + lim + lim r →r0 r →r0 r →r0 y − y0 y − y0 y − y0 1444 24444 4 3 =− ∂f ∂y f ( x, y ) − f ( x 0 , y 0 ) f ( x 0 , y 0 ) − f ( x, y 0 ) x − x0 ∂f = lim + lim lim − ( r0 ) r →r0 r →r0 r →r0 y − y y − y0 x − x0 ∂y 0 1444 24444 4 3 =− ∂f ∂x
- 16. Old differentiability ⇒ new differentiability f ( x, y ) − f ( x0 , y 0 ) ∂f x − x0 ∂f = lim − − ( r0 ) rlim ( r0 ) r →r0 →r0 y − y y − y0 ∂x ∂y 1 3 2 1 24 0 4 3 h sinθ =b / a
- 17. Old differentiability ⇒ new differentiability f ( x, y ) − f ( x0 , y 0 ) b ∂f ∂f = lim − ( r0 ) − ( r0 ) r →r0 h sinθ a ∂x ∂y
- 18. Old differentiability ⇒ new differentiability = f ( x0 + bh, y 0 + ah ) − f ( x0 , y 0 ) b ∂f 1 ∂f lim − r0 ) − ( ( r0 ) h→0 sinθ 1444444 h ∂y 2444444 3 a ∂x Duf u = ( b, a )
- 19. Old differentiability ⇒ new differentiability 1 b ∂f ∂f = Duf ( r0 ) − ( r0 ) − ( r0 ) a a ∂x ∂y u = ( b, a )
- 20. Thm: If f is old differentiable at r0 then Duf(r0) exists and: Duf ( r0 ) ∂f ∂f = ( r0 ) b + ( r0 ) a ∂x ∂y = ∇f ( r0 ) ×u • Stewart thm. 11.6.3, Leithold thm. 12.6.2 • u = (b, a)
- 21. Old differentiability ⇒ new differentiability f ( x, y ) − f ( x, y 0 ) ∂f lim − ( r0 ) r →r0 y − y0 ∂y b ∂f 1 ∂f ∂f ∂f = b ( r0 ) + a ( r0 ) − ( r0 ) − ( r0 ) a ∂x ∂y ∂y a ∂x = 0.
- 22. Conclusion • Old differentiability ⇒ new differentiability ⇒ chain rule.
- 23. Question: • Old differentiability ⇔ new differentiability?
- 24. References • Stewart, Essential Calculus (Thompson, 2007) • Leithold, The Calculus 7th ed. (Harper, 1996)