Question

nth differentiation of x3 cosx

Answer 1

Answer:

Step-by-step explanation:

-3sinx

Answer 2

SOLUTION

TO DETERMINE

The nth derivative of x³ cos x

EVALUATION

Here the given function is

f(x) = x³ cos x

Let u = cos x and v = x³

Then

$\displaystyle \sf{u_n = \cos \bigg( \frac{n\pi}{2} + x \bigg)}$

$\sf{v_1 = 3 {x}^{2} }$

$\sf{v_2 = 6x }$

$\sf{v_3 = 6 }$

$\sf{v_n= 0 \: \: \: for \: \: n > 3 }$

Thus the function can be rewritten as

f(x) = uv

Differentiating both sides n times with respect to x using Leibnitz theorem we get

$\sf{ {f}^{n}(x) }$

$\sf{ = (uv)_n}$

$\displaystyle \sf{ = {}^{n} C_0u_{n}v +{}^{n} C_1u_{n - 1}v_1 +{}^{n} C_2u_{n - 2}v_2 +{}^{n} C_3u_{n}v_3 + {}^{n} C_4u_{n - 4}v_4 + .. \: .. }$

$\displaystyle \sf = {}^{n} C_0\cos \bigg( \frac{n\pi}{2} + x \bigg) {x}^{3} +{}^{n} C_1\cos \bigg( \frac{(n - 1)\pi}{2} + x \bigg)3 {x}^{2} +{}^{n} C_2\cos \bigg( \frac{(n - 2)\pi}{2} + x \bigg)6x +{}^{n} C_3\cos \bigg( \frac{(n - 3)\pi}{2} + x \bigg)6 +{}^{n} C_4\cos \bigg( \frac{(n - 4)\pi}{2} + x \bigg).0 + .. \: ..$

$\displaystyle \sf = {}^{n} C_0\cos \bigg( \frac{n\pi}{2} + x \bigg) {x}^{3} +{}^{n} C_1\cos \bigg( \frac{(n - 1)\pi}{2} + x \bigg)3 {x}^{2} +{}^{n} C_2\cos \bigg( \frac{(n - 2)\pi}{2} + x \bigg)6x +{}^{n} C_3\cos \bigg( \frac{(n - 3)\pi}{2} + x \bigg)6$

$\displaystyle \sf = {x}^{3} \cos \bigg( \frac{n\pi}{2} + x \bigg) +3n {x}^{2} \cos \bigg( \frac{(n - 1)\pi}{2} + x \bigg) +3n(n - 1)x\cos \bigg( \frac{(n - 2)\pi}{2} + x \bigg) +n(n - 1)(n - 2)\cos \bigg( \frac{(n - 3)\pi}{2} + x \bigg)$

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