Successive Differentiation – Leibnitz’s Theorem

Leibnitz’s Theorem works on finding successive derivatives of product of two derivable functions.

Statement:

If u and v are two functions of x, each possessing derivatives upto n^th order, then the product y=u.v is derivable n times and
\[{{y}_{n}}={{\left( u.v \right)}_{n}}={}^{n}{{C}_{0}}{{u}_{n}}v+{}^{n}{{C}_{1}}{{u}_{n-1}}{{v}_{1}}+{}^{n}{{C}_{2}}{{u}_{n-2}}{{v}_{2}}+…+{}^{n}{{C}_{r}}{{u}_{n-r}}{{v}_{r}}+…+{}^{n}{{C}_{n}}u{{v}_{n}}\]
\[\Rightarrow {{y}_{n}}={{u}_{n}}v+{}^{n}{{C}_{1}}{{u}_{n-1}}{{v}_{1}}+{}^{n}{{C}_{2}}{{u}_{n-2}}{{v}_{2}}+…+{}^{n}{{C}_{r}}{{u}_{n-r}}{{v}_{r}}+…+u{{v}_{n}}\]
\[As,{}^{n}{{C}_{r}}=\frac{n!}{r!(n-r)!}\Rightarrow {}^{n}{{C}_{0}}={}^{n}{{C}_{n}}=1\]
Where y_r, u_r, v_r denote the r^th derivatives of y, u and v respectively with respect to x.
Proof:
Since y = u.v, differentiating both sides w.r.t. x we get,
\[{{y}_{1}}={{u}_{1}}v+u{{v}_{1}}\]
\[\therefore {{y}_{2}}={{u}_{2}}v+{{u}_{1}}{{v}_{1}}+{{u}_{1}}{{v}_{1}}+u{{v}_{2}}={{u}_{2}}v+2{{u}_{1}}{{v}_{1}}+u{{v}_{2}}\]
\[\Rightarrow {{y}_{2}}={{u}_{2}}v+{}^{2}{{C}_{1}}{{u}_{1}}{{v}_{1}}+u{{v}_{2}}\]
Thus the theorem is found to be true for n = 1, 2.
Let us assume that the theorem is true for a particular positive integral value of n, say, k, where k < n.
Then we have,
\[{{y}_{k}}={{u}_{k}}v+{}^{k}{{C}_{1}}{{u}_{k-1}}{{v}_{1}}+{}^{k}{{C}_{2}}{{u}_{k-2}}{{v}_{2}}+…+{}^{k}{{C}_{r-1}}{{u}_{k-r+1}}{{v}_{r-1}}+{}^{k}{{C}_{r}}{{u}_{k-r}}{{v}_{r}}+…+u{{v}_{k}}\]
Differentiating both sides again w.r.t. x we get,
\[{{y}_{k+1}}={{u}_{k+1}}v+{{u}_{k}}{{v}_{1}}+{}^{k}{{C}_{1}}\left( {{u}_{k}}{{v}_{1}}+{{u}_{k-1}}{{v}_{2}} \right)\]
\[+{}^{k}{{C}_{2}}\left( {{u}_{k-1}}{{v}_{2}}+{{u}_{k-2}}{{v}_{3}} \right)+…+{}^{k}{{C}_{r-1}}\left( {{u}_{k-r+2}}{{v}_{r-1}}+{{u}_{k-r+1}}{{v}_{r}} \right)\]
\[+{}^{k}{{C}_{r}}\left( {{u}_{k-r+1}}{{v}_{r}}+{{u}_{k-r}}{{v}_{r+1}} \right)+…+{{u}_{1}}{{v}_{k}}+u{{v}_{k+1}}\]
\[\Rightarrow {{y}_{k+1}}={{u}_{k+1}}v+\left( 1+{}^{k}{{C}_{1}} \right){{u}_{k}}{{v}_{1}}+\left( {}^{k}{{C}_{1}}+{}^{k}{{C}_{2}} \right){{u}_{k-1}}{{v}_{2}}+…+\]
\[\left( {}^{k}{{C}_{r-1}}+{}^{k}{{C}_{r}} \right){{u}_{k-r+1}}{{v}_{r}}+…+u{{v}_{k+1}}\]
We know that,
\[{}^{k}{{C}_{r-1}}+{}^{k}{{C}_{r}}={}^{k+1}{{C}_{r}},{}^{k}{{C}_{1}}+{}^{k}{{C}_{2}}={}^{k+1}{{C}_{2}},1+{}^{k}{{C}_{1}}={}^{k+1}{{C}_{1}}\]
\[\therefore {{y}_{k+1}}={{u}_{k+1}}v+{}^{k+1}{{C}_{1}}{{u}_{k}}{{v}_{1}}+{}^{k+1}{{C}_{2}}{{u}_{k-1}}{{v}_{2}}+…\]
\[+{}^{k+1}{{C}_{r}}{{u}_{k-r+1}}{{v}_{r}}+…+u{{v}_{k+1}}\]
Thus we find that if the theorem is assumed to be true for any positive integral value k of n, then it is found to be true for the next integral value (k+1) of n.
And the theorem has already been found to be true for n =1, 2.
Hence, by the principle of Mathematical Induction, the theorem is true for every positive integral value of n.
Thus Leibnitz’s Theorem is established.
Now is the time to check some problems to find the n^th order derivative using Leibnitz’s Theorem.

Note: Generally you should initialize v with the function which will take minimum iterations to become 0.

\[If,y={{e}^{x}}{{x}^{3}}find,{{y}_{4}}\]
Solution:
\[Let,u={{e}^{x}},v={{x}^{3}}\]
Then by Leibnitz’s Theorem,
\[{{y}_{4}}={{u}_{4}}v+{}^{4}{{C}_{1}}{{u}_{3}}{{v}_{1}}+{}^{4}{{C}_{2}}{{u}_{2}}{{v}_{2}}+{}^{4}{{C}_{3}}{{u}_{1}}{{v}_{3}}+u{{v}_{4}}\]
\[\Rightarrow {{y}_{4}}={{D}^{4}}({{e}^{x}}).{{x}^{3}}+{}^{4}{{C}_{1}}{{D}^{3}}({{e}^{x}}).D\left( {{x}^{3}} \right)+{}^{4}{{C}_{2}}{{D}^{2}}({{e}^{x}}).{{D}^{2}}\left( {{x}^{3}} \right)\]
\[+{}^{4}{{C}_{3}}D({{e}^{x}}).{{D}^{3}}\left( {{x}^{3}} \right)+{{e}^{x}}.{{D}^{4}}\left( {{x}^{3}} \right),where,{{D}^{r}}\equiv \frac{{{d}^{r}}}{d{{x}^{r}}}\]
\[\Rightarrow {{y}_{4}}={{e}^{x}}{{x}^{3}}+4{{e}^{x}}.3{{x}^{2}}+6.{{e}^{x}}.6x+4{{e}^{x}}.6+{{e}^{x}}.0\]
\[\therefore {{y}_{4}}={{e}^{x}}\left\{ {{x}^{3}}+12{{e}^{x}}{{x}^{2}}+36.{{e}^{x}}x+24 \right\}\]

\[If,y=\sin \left( m{{\sin }^{-1}}x \right),show-that\]
\[(i)\left( 1-{{x}^{2}} \right){{y}_{2}}-x{{y}_{1}}+{{m}^{2}}y=0\]
\[(ii)\left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+1 \right)x{{y}_{n+1}}+\left( {{m}^{2}}-{{n}^{2}} \right){{y}_{n}}=0\]
Solution:
\[Given,y=\sin \left( m{{\sin }^{-1}}x \right)\]
\[\therefore {{y}_{1}}=\cos \left( m{{\sin }^{-1}}x \right).m.\frac{1}{\sqrt{1-{{x}^{2}}}}\]
\[\Rightarrow \sqrt{1-{{x}^{2}}}.{{y}_{1}}=m\cos \left( m{{\sin }^{-1}}x \right)\]
\[\Rightarrow \left( 1-{{x}^{2}} \right).{{y}_{1}}^{2}={{m}^{2}}{{\cos }^{2}}\left( m{{\sin }^{-1}}x \right)={{m}^{2}}\left\{ 1-{{\sin }^{2}}\left( m{{\sin }^{-1}}x \right) \right\}\]
\[\Rightarrow \left( 1-{{x}^{2}} \right).{{y}_{1}}^{2}={{m}^{2}}\left( 1-{{y}^{2}} \right)\]
Differentiating both sides w.r.t. x we get,
\[\left( 1-{{x}^{2}} \right).2{{y}_{1}}.{{y}_{2}}-2x.{{y}_{1}}^{2}=-2{{m}^{2}}y{{y}_{1}}\]
\[\therefore \left( 1-{{x}^{2}} \right){{y}_{2}}-x{{y}_{1}}+{{m}^{2}}y=0,[\because {{y}_{1}}\ne 0]\]
Now applying Leibnitz’s theorem to differentiate the above equation n times w.r.t. x we get,
\[{{D}^{n}}\left\{ \left( 1-{{x}^{2}} \right){{y}_{2}} \right\}-{{D}^{n}}\left\{ x{{y}_{1}} \right\}+{{m}^{2}}{{y}_{n}}=0\]
\[\Rightarrow \left\{ \left( 1-{{x}^{2}} \right){{y}_{n+2}}+{}^{n}{{C}_{1}}.{{y}_{n+1}}.(-2x)+{}^{n}{{C}_{2}}.{{y}_{n}}.(-2)+0 \right\}\]
\[-\left\{ x{{y}_{n+1}}+{}^{n}{{C}_{1}}.{{y}_{n}}.1+0 \right\}+{{m}^{2}}{{y}_{n}}=0\]
\[\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{n+2}}-2nx{{y}_{n+1}}-n(n-1){{y}_{n}}-x{{y}_{n+1}}-n{{y}_{n}}+{{m}^{2}}{{y}_{n}}=0\]
\[\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+1 \right)x{{y}_{n+1}}-\left( {{n}^{2}}-n+n \right){{y}_{n}}+{{m}^{2}}{{y}_{n}}=0\]
\[\therefore \left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+1 \right)x{{y}_{n+1}}+\left( {{m}^{2}}-{{n}^{2}} \right){{y}_{n}}=0\]

\[If,y={{e}^{m{{\sin }^{-1}}x}},show-that\]
\[(i)\left( 1-{{x}^{2}} \right){{y}_{2}}-x{{y}_{1}}-{{m}^{2}}y=0\]
\[(ii)\left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+1 \right)x{{y}_{n+1}}-\left( {{m}^{2}}+{{n}^{2}} \right){{y}_{n}}=0\]
\[Also,find\to {{y}_{n}},when\to x=0\]
Solution:
\[Given,y={{e}^{m{{\sin }^{-1}}x}}\]
\[\therefore {{y}_{1}}={{e}^{m{{\sin }^{-1}}x}}.m.\frac{1}{\sqrt{1-{{x}^{2}}}}………..(1)\]
\[\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{1}}^{2}={{m}^{2}}{{\left( {{e}^{m{{\sin }^{-1}}x}} \right)}^{2}}={{m}^{2}}{{y}^{2}}\]
\[\therefore \left( 1-{{x}^{2}} \right).2{{y}_{1}}.{{y}_{2}}-2x{{y}_{1}}^{2}=2{{m}^{2}}y{{y}_{1}}\]
\[\therefore \left( 1-{{x}^{2}} \right){{y}_{2}}-x{{y}_{1}}-{{m}^{2}}y=0………..(2)\]
Now applying Leibnitz’s theorem to differentiate the above equation n times w.r.t. x we get,
\[{{D}^{n}}\left\{ \left( 1-{{x}^{2}} \right){{y}_{2}} \right\}-{{D}^{n}}\left\{ x{{y}_{1}} \right\}-{{m}^{2}}{{y}_{n}}=0\]
\[\Rightarrow \left\{ \left( 1-{{x}^{2}} \right){{y}_{n+2}}+{}^{n}{{C}_{1}}.{{y}_{n+1}}.(-2x)+{}^{n}{{C}_{2}}.{{y}_{n}}.(-2)+0 \right\}\]
\[-\left\{ x{{y}_{n+1}}+{}^{n}{{C}_{1}}.{{y}_{n}}.1+0 \right\}-{{m}^{2}}{{y}_{n}}=0\]
\[\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{n+2}}-2nx{{y}_{n+1}}-n(n-1){{y}_{n}}-x{{y}_{n+1}}-n{{y}_{n}}-{{m}^{2}}{{y}_{n}}=0\]
\[\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+1 \right)x{{y}_{n+1}}-\left( {{n}^{2}}-n+n \right){{y}_{n}}-{{m}^{2}}{{y}_{n}}=0\]
\[\therefore \left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+1 \right)x{{y}_{n+1}}-\left( {{m}^{2}}+{{n}^{2}} \right){{y}_{n}}=0……….(3)\]
\[\because y={{e}^{m{{\sin }^{-1}}x}}\Rightarrow y=1,when,x=0\]
Also from (1), (2) and (3)
\[{{y}_{1}}=m,{{y}_{2}}={{m}^{2}},{{y}_{n+2}}=\left( {{m}^{2}}+{{n}^{2}} \right){{y}_{n}}\]
Now putting n = 1, 2, 3, … in the relation
\[{{y}_{n+2}}=\left( {{m}^{2}}+{{n}^{2}} \right){{y}_{n}},we-have\]
\[{{y}_{3}}=\left( {{m}^{2}}+{{1}^{2}} \right){{y}_{1}}=m\left( {{m}^{2}}+{{1}^{2}} \right)\]
\[{{y}_{4}}=\left( {{m}^{2}}+{{2}^{2}} \right){{y}_{2}}=m\left( {{m}^{2}}+{{2}^{2}} \right)\]
\[{{y}_{5}}=\left( {{m}^{2}}+{{3}^{2}} \right){{y}_{3}}=m\left( {{m}^{2}}+{{1}^{2}} \right)\left( {{m}^{2}}+{{3}^{2}} \right)\]
\[{{y}_{6}}=\left( {{m}^{2}}+{{4}^{2}} \right){{y}_{4}}=m\left( {{m}^{2}}+{{2}^{2}} \right)\left( {{m}^{2}}+{{4}^{2}} \right)\]
Hence when x = 0
\[{{y}_{n}}=m\left( {{m}^{2}}+{{1}^{2}} \right)\left( {{m}^{2}}+{{3}^{2}} \right)…\left\{ {{m}^{2}}+{{\left( n-2 \right)}^{2}} \right\}if\to ‘n’is\_odd\]
\[{{y}_{n}}=m\left( {{m}^{2}}+{{2}^{2}} \right)\left( {{m}^{2}}+{{4}^{2}} \right)…\left\{ {{m}^{2}}+{{\left( n-2 \right)}^{2}} \right\}if\to ‘n’is\_even\]

\[If,f(x)={{x}^{n}},prove-that\]
\[f(1)+\frac{f'(1)}{1!}+\frac{f”(1)}{2!}+\frac{f”'(1)}{3!}+…+\frac{{{f}^{n}}(1)}{n!}={{2}^{n}}\]
Solution:
\[Given,f(x)={{x}^{n}}\]
\[\therefore f'(x)=n{{x}^{n-1}},f”(x)=n(n-1){{x}^{n-2}}…\]
\[\therefore {{f}^{r}}(x)=n(n-1)(n-2)…(n-r+1){{x}^{n-r}},\left( r\le n \right)\]
\[\therefore {{f}^{n}}(x)=n!\]
\[L.H.S.=f(1)+\frac{f'(1)}{1!}+\frac{f”(1)}{2!}+\frac{f”'(1)}{3!}+…+\frac{{{f}^{n}}(1)}{n!}\]
\[=1+\frac{n}{1!}+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!}+…….+\frac{n!}{n!}\]
\[={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+{}^{n}{{C}_{3}}+….+{}^{n}{{C}_{n}}={{2}^{n}}=R.H.S.\]

\[If,y=\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}},|x|<1,show-that\]
\[\left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+3 \right)x{{y}_{n+1}}-{{\left( n+1 \right)}^{2}}{{y}_{n}}=0\]
Solution:
\[Given,y=\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}},|x|<1\]
\[\Rightarrow \left( 1-{{x}^{2}} \right){{y}^{2}}={{\left( {{\sin }^{-1}}x \right)}^{2}}\]
Now, differentiating w.r.t. x we get,
\[\left( 1-{{x}^{2}} \right).2y{{y}_{1}}-2x{{y}^{2}}=2{{\sin }^{-1}}x.\frac{1}{\sqrt{1-{{x}^{2}}}}=2y\]
\[\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{1}}-xy=1,assu\min g,y\ne 0\]
Again, differentiating w.r.t. x we get,
\[\left( 1-{{x}^{2}} \right){{y}_{2}}-2x{{y}_{1}}-x{{y}_{1}}-1.y=0\]
\[\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{2}}-3x{{y}_{1}}-y=0……….(1)\]
Now applying Leibnitz’s theorem to differentiate the above equation n times w.r.t. x we get,
\[{{D}^{n}}\left\{ \left( 1-{{x}^{2}} \right){{y}_{2}} \right\}-{{D}^{n}}\left\{ 3x{{y}_{1}} \right\}-{{y}_{n}}=0\]
\[\Rightarrow \left\{ \left( 1-{{x}^{2}} \right){{y}_{n+2}}+{}^{n}{{C}_{1}}.{{y}_{n+1}}.(-2x)+{}^{n}{{C}_{2}}.{{y}_{n}}.(-2)+0 \right\}\]
\[-3\left\{ x{{y}_{n+1}}+{}^{n}{{C}_{1}}.{{y}_{n}}.1+0 \right\}-{{y}_{n}}=0\]
\[\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{n+2}}-2nx{{y}_{n+1}}-n(n-1){{y}_{n}}-3x{{y}_{n+1}}-3n{{y}_{n}}-{{y}_{n}}=0\]
\[\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+3 \right)x{{y}_{n+1}}-{{\left( n+1 \right)}^{2}}{{y}_{n}}=0\]

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