Find the nth differential coefficient of x^3 cos x

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Find the nth differential coefficient of  x^3 cos x . This solution employs Leibnitz’s theorem, addressing the application and simplification of the nth derivative. For x^3 cos x , higher-order derivatives of x^3  beyond the fourth will become zero, simplifying the calculations. The detailed solution explores the use of trigonometric identities and combinations to derive the nth differential coefficient, resulting in a comprehensive expression involving both cosine and sine terms. This step-by-step approach provides clarity on advanced calculus concepts and their applications in differential equations.

Q 1:- Find the nth differential coefficient of x^3\cos x.

Solution:-  Since, the fourth and higher derivatives of x³ will become zero, therefore for the sake of convenience we should choose x³ as the second function. Applying Leibnitz’s theorem, we have

D^n(x^3\cos x) = {}^nC_0(D^n\cos x).x^3 + {}^nC_1(D^{n-1}\cos x)(Dx^3) + {}^nC_2(D^{n-2}\cos x)(D^2x^3) + {}^nC_3(D^{n-3}\cos x)(D^3x^3)

We know that, D^n\cos x =\cos (x + n\pi/2)

D^n(x^3\cos x) = 1.\cos (x + n\pi/2).x^3 + n.\cos [x + (n-1)\pi/2]3x^2 + \dfrac{n(n-1)}{1.2}.\cos [x + (n-2)\pi/2].6x + \dfrac{n(n-1)(n-2)}{1.2.3}\cos [x + (n-3)\pi/2]6

Since, \cos [x + (n-1)\pi/2] = \sin(x + n\pi/2), \cos [x + (n-2)\pi/2] = -\cos[x + n\pi/2]

And \cos [x + (n-3)\pi/2] = -\sin[x +n\pi/2]

\Rightarrow D^n(x^3\cos x) = x^3\cos (x + n\pi/2) + 3nx^2\sin(x + n\pi/2) - 3n(n-1)x\cos(x + n\pi/2) - n(n-1)(n-2)\sin(x +n\pi/2)

\Rightarrow D^n(x^3\cos x) = [x^3 - 3n(n-1)x]\cos(x + n\pi/2)+[3x^2n - n(n-1)(n-2)]\sin(x + n\pi/2).

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