Successive Differentiation


Differential Calculus / Wednesday, August 22nd, 2018
(Last Updated On: August 22, 2018)

Finding nth Order Derivative

Successive Differentiation – The derivative or the differential coefficient of a function ‘f’ of a single variable has already been defined in my previous articles.

  1. First Order Derivative Solution – Part I
  2. First Order Derivative Solution – Part II
  3. Second Order Derivative Solution

It has been found that the derivative or the differential coefficient f’(x) of a function f of a single variable x defined in a domain D is again another function of x defined in a certain domain D’ which is generally a subset of D.

Now it may so happen that the derived function f’(x) can be differentiated again with respect to the variable x to give another function of x. This new function, denoted by f’’(x), is called second order derivative of the function f. if this function f’’(x) can again be differentiated with respect to x, we shall get the third order derivative, denoted by f’’’(x). This process of differentiating a function successively n times will give fn(x), where n is any positive integer, provided the function admits of nth derivative.

The use and importance of successive derivative of a function will be appreciated in the discussion on expansion of functions in series and formation of differential equations.

Example:
\[If,y=f(x)=7{{x}^{3}},then\]
\[Dy=\frac{dy}{dx}={{y}_{1}}=f'(x)=21{{x}^{2}}\]
\[{{D}^{2}}y=\frac{{{d}^{2}}y}{d{{x}^{2}}}={{y}_{2}}=f”(x)=42x\]
\[{{D}^{3}}y=\frac{{{d}^{3}}y}{d{{x}^{3}}}={{y}_{3}}=f”'(x)=42\]
\[{{D}^{4}}y=\frac{{{d}^{4}}y}{d{{x}^{4}}}={{y}_{4}}={{f}^{iv}}(x)=0\]
\[and,{{y}_{5}}={{y}_{6}}={{y}_{7}}=…=0\]

n-th derivative of some Particular Functions

Let m, n are positive integer such that m > n.
\[(i)If,y={{x}^{m}}(x>0)\Rightarrow {{y}_{n}}=m(m-1)(m-2)…(m-n+1){{x}^{m-n}}\]
\[(ii)If,y={{(ax+b)}^{m}},where,a,b-cons\tan t\]
\[\Rightarrow {{y}_{n}}=m(m-1)(m-2)…(m-n+1).{{a}^{n}}{{(ax+b)}^{m-n}}\]
\[\therefore {{D}^{n}}{{(ax+b)}^{m}}=\frac{m!}{\left( m-n \right)!}{{a}^{n}}{{(ax+b)}^{m-n}}\]
\[(iii){{D}^{n}}({{e}^{ax}})={{a}^{n}}{{e}^{ax}}\]
\[(iv){{D}^{n}}({{a}^{x}})={{\left( {{\log }_{e}}a \right)}^{n}}.{{a}^{x}}\]
\[(v){{D}^{n}}\left( \frac{1}{x+a} \right)=\frac{{{\left( -1 \right)}^{n}}.n!}{{{\left( x+a \right)}^{n+1}}}\]
\[(vi){{D}^{n}}\left\{ \frac{1}{{{(ax+b)}^{m}}} \right\}=\frac{{{\left( -1 \right)}^{n}}.{{a}^{n}}.(m+n-1)!}{(m-1)!{{\left( ax+b \right)}^{m+n}}}\]
\[(vii){{D}^{n}}\left\{ {{\log }_{e}}(x+a) \right\}=\frac{{{\left( -1 \right)}^{n-1}}(n-1)!}{{{\left( x+a \right)}^{n}}}\]
\[(viii){{D}^{n}}\left\{ sin(ax+b) \right\}={{a}^{n}}\sin \left( n.\frac{\pi }{2}+ax+b \right)\]
\[(ix){{D}^{n}}\left\{ \cos (ax+b) \right\}={{a}^{n}}\cos \left( n.\frac{\pi }{2}+ax+b \right)\]

Now we will solve some problems to understand the topic.

 Question 01

\[If,y=\frac{x}{x+1},show-that,{{y}_{5}}(0)=5!\]

Solution:
\[Here,y=\frac{x}{x+1}=\frac{x+1-1}{x+1}=1-\frac{1}{x+1}=1-{{\left( x+1 \right)}^{-1}}\]
\[\therefore {{y}_{1}}=(-1)(-1){{\left( x+1 \right)}^{-2}}\]
\[\therefore {{y}_{2}}=(-1)(-1)(-2){{\left( x+1 \right)}^{-3}}={{(-1)}^{3}}.2!{{\left( x+1 \right)}^{-3}}\]
\[\therefore {{y}_{3}}={{(-1)}^{4}}.3!{{\left( x+1 \right)}^{-4}}\]
Proceeding in this way
\[{{y}_{5}}={{(-1)}^{6}}.5!{{\left( x+1 \right)}^{-6}}\]
\[\therefore {{y}_{5}}(0)={{(-1)}^{6}}.5!{{\left( 0+1 \right)}^{-6}}=5!\]

 Question 02

\[If,y=\frac{{{x}^{2}}+1}{(x-1)(x-2)(x-3)},find,{{y}_{n}}\]
Solution:
Let us first decompose y in partial fractions.
\[Let,\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}=\frac{{{x}^{2}}+1}{(x-1)(x-2)(x-3)}\]
Multiplying both sides by (x-1)(x-2)(x-3), we get,
\[A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)={{x}^{2}}+1……(1)\]
Putting x=1.2 and 3 successively on both sides of (1) we have, A = 1, B = -5, C = 5
\[Thus,y=\frac{1}{x-1}-\frac{5}{x-2}+\frac{5}{x-3}\]
\[\therefore {{y}_{n}}=\frac{{{(-1)}^{n}}n!}{{{(x-1)}^{n+1}}}-5.\frac{{{(-1)}^{n}}n!}{{{(x-2)}^{n+1}}}+5.\frac{{{(-1)}^{n}}n!}{{{(x-3)}^{n+1}}}\]
\[\therefore {{y}_{n}}={{(-1)}^{n}}n!\left\{ \frac{1}{{{(x-1)}^{n+1}}}-\frac{5}{{{(x-2)}^{n+1}}}+\frac{5}{{{(x-3)}^{n+1}}} \right\}\]

 Question 03

\[If,y=\frac{1}{{{x}^{2}}+{{y}^{2}}},show-that,{{y}_{n}}=\frac{{{(-1)}^{n}}n!\sin (n+1)\theta }{a{{r}^{n+1}}}\]
\[where,r=\sqrt{{{a}^{2}}+{{x}^{2}}},and,r\cos \theta =x;r\sin \theta =a,-\pi <\theta \le \pi \]
Solution:
\[y=\frac{1}{{{x}^{2}}+{{y}^{2}}}=\frac{1}{(x+ia)(x-ia)}=\frac{1}{2ia}\left\{ \frac{1}{x-ia}-\frac{1}{x+ia} \right\}[\because {{i}^{2}}=-1]\]
\[\therefore {{y}_{n}}=\frac{1}{2ia}\left\{ \frac{{{(-1)}^{n}}n!}{{{\left( x-ia \right)}^{n+1}}}-\frac{{{(-1)}^{n}}n!}{{{\left( x+ia \right)}^{n+1}}} \right\}\]
\[\Rightarrow {{y}_{n}}=\frac{{{(-1)}^{n}}n!}{2ia}\left\{ {{\left( x-ia \right)}^{-n-1}}-{{\left( x+ia \right)}^{-n-1}} \right\}……….(1)\]
Let us put x = r cosθ and a = rsinθ
\[Then,{{a}^{2}}+{{x}^{2}}={{r}^{2}},and,x-ia=r(\cos \theta -i\sin \theta )\]
\[\therefore {{\left( x-ia \right)}^{-n-1}}={{r}^{-(n+1)}}{{(\cos \theta -i\sin \theta )}^{-(n+1)}}\]
By De Moivre’s theorem,
\[\Rightarrow {{\left( x-ia \right)}^{-n-1}}={{r}^{-(n+1)}}\left\{ \cos (n+1)\theta +i\sin (n+1)\theta  \right\}\]
Similarly,
\[{{\left( x+ia \right)}^{-n-1}}={{r}^{-(n+1)}}\left\{ \cos (n+1)\theta -i\sin (n+1)\theta  \right\}\]
Using these values we get from (1)
\[{{y}_{n}}=\frac{{{(-1)}^{n}}n!}{2ia}.{{r}^{-(n+1)}}\left\{ \cos (n+1)\theta +i\sin (n+1)\theta -\cos (n+1)\theta +i\sin (n+1)\theta  \right\}\]
\[\therefore {{y}_{n}}=\frac{{{(-1)}^{n}}n!\sin (n+1)\theta }{a{{r}^{n+1}}}\]

 Question 04

\[If,y={{e}^{ax}}\cos bx,show-that,{{y}_{n}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{\frac{n}{2}}}{{e}^{ax}}\cos (bx+n\theta )\]
\[where,\cos \theta =\frac{a}{r},\sin \theta =\frac{b}{r},and,{{r}^{2}}={{a}^{2}}+{{b}^{2}}.-\pi <\theta \le \pi \]
Solution:
\[Given,y={{e}^{ax}}\cos bx\]
\[\therefore {{y}_{1}}=a{{e}^{ax}}\cos bx-b{{e}^{ax}}\sin bx={{e}^{ax}}\left( a\cos bx-b\sin bx \right)\]
We can find two real numbers r and θ, such that, rcosθ = a, rsinθ = b;
\[then,r=+\sqrt{{{a}^{2}}+{{b}^{2}}}and,\cos \theta =\frac{a}{r},\sin \theta =\frac{b}{r},where-\pi <\theta \le \pi \]
With these substitutions, we have
\[{{y}_{1}}={{e}^{ax}}.r\left( \cos bx.\cos \theta -\sin bx.\sin \theta  \right)={{e}^{ax}}.r\cos (bx+\theta )\]
Similarly,
\[{{y}_{2}}={{e}^{ax}}.r.r\cos (bx+\theta +\theta )={{e}^{ax}}.{{r}^{2}}\cos (bx+2\theta )\]
\[\therefore {{y}_{n}}={{e}^{ax}}.{{r}^{n}}\cos (bx+n\theta )={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{\frac{n}{2}}}{{e}^{ax}}\cos (bx+n\theta )\]
\[where,\cos \theta =\frac{a}{r},\sin \theta =\frac{b}{r},and,{{r}^{2}}={{a}^{2}}+{{b}^{2}}.-\pi <\theta \le \pi \] 

 

In the next article we will learn Leibnitz Theorem of Successive Differentiation.

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