Leibnitz Theorem

Leibnitz Theorem

Leibnitz a German philosopher and mathematician proposed a rule which is a generalization of  the product rule of differentiation. Leibnitz rule states that if two functions u(x) and v(x) are differentiable n times individually, then their product u(x).v(x) is also differentiable n times. 

If u(x) and v(x) are two n-times differentiable functions then

 

(uv)n=unv+nC1un–1v1+nC2un–2v2+⋯+nCrun–rvr+⋯+uvn

Proof Of Leibnitz Theorem 

Leibnitz formula can be proved by Principle of Mathematical Induction.  

Let us start with n =1 

u(x).v(x))' = u'(x).v(x) + u(x).v'(x)

for second derivative n=2

(u(x).v(x))'' = u''(x).v(x) + 2u'(x).v'(x) + u(x).v''(x)

for third derivative

(uv)′′′  = = (u”v+2(u′v′)+uv”)′

= u′′′v+u′′v′+2u′′v′+2u′v′′+u′v′′+uv′′′

= u′′′v+3u′′v′+3u′v′′+uv′′

We know 

dⁿ(UV)/dxⁿ = nC₀.U_n.V + nC₁.U_n-1.V₁ + nC₂.U_n-₂.V₂ + nC_r.U_n-r.V_r +...........+ nC_n.U.V_n

We assume that the theorem is true for a particular value of n say k

(UV)k = Uk.V + kC1 Uk−1.V1 + kC2 Uk−2.V2 + ...+ kCr-1Uk−r+1.Vr−1 + kCr Uk−r.Vr + kCkU.Vk

Differentiating both sides, we get 

(UV)k+1 = [Uk+1.V + Uk.V1] + [kC1 Uk.V1 + kC1 . Uk−1.V2] +...+ [kC2 Uk−1.V2 + kC2 Uk−2.V3] +

...+ [kCr−1 Uk−r+2.Vr−1] + [kCr−1 Uk−r+1.Vr] + [kCr Uk−r+1.Vr + kCr Uk−r.Vr+1] +...

+ [kCk U1.Vk + kCk U.Vk+1 ]

Re-arranging the terms

 =[Uk+1.V +(1 + kC1) Uk.V1] + (kC1 + kC2) Uk−1.V2 +...+( kCr−1+ kCr ) Uk−r+1.Vr +... kCk U.Vk+1

We know kCk = 1 ; (1 + kC1) = ( kCk + kC1)

Also kCr−1 + kCr = k+1 Cr

  =Uk+1.V + (k+1)C1 Uk.V1 + (k+1)C2Uk−1.V2 +...+ (k+1)Cr Uk−r+1.Vr +...+ U.Vk+1 

Thus, the theorem is true for n =1,2,3 and for n = k+1, Hence, the theorem is true for all positive real values of n.

Proof  : Method 2

If two functions f(x) and g(x) are differentiable n times separately, then their product f(x). g(x) is also differentiable n times. Let us generalise the Leibniz rule using the formula 

           

dndxn(fg)=∑i=0n(ni)f(i)g(n−i)

 

let assume that this true for  1< m < n

dmdxm(fg)=∑i=0m(mi)f(m)g(m−i)

 

dm+1dxm+1(fg)=dmdxm(ddx(fg))=dmdxm(f′g)+dmdxm(fg′)

  




g(m−i)

 dm+1 
=∑i=0m(mi)f(i+1)g(m−i)+∑i=0m(mi)f(i)g(m+1−i)=f(m+1)g+∑i=1m(mi−1)f(i)g(m+1−i)+∑i=1m(mi)f(i)g(m+1−i)+fg(m+1)=f(m+1)g+∑i=1m((mi−1)+(mi))f(i)g(m+1−i)+fg(m+1)=∑i=0m+1(m+1i)f(i)g(m+1−i)

applying Leibniz theorem

dm+1dxm+1(fg)=∑i=0m(mi)f(i+1)g(m−i)+∑i=0m(mi)f(i)g(m+1−i)=f(m+1)g+∑i=1m(mi−1)f(i)g(m+1−i)+∑i=1m(mi)f(i)g(m+1−i)+fg(m+1)=f(m+1)g+∑i=1m((mi−1)+(mi))f(i)g(m+1−i)+fg(m+1)=∑i=0m+1(m+1i)f(i)g(m+1−i).

(fg)=∑i=0m(mi)f(i+1)g(m−i)+∑i=0m(mi)f(i)g(m+1−i)