Find the nth derivative of x3 cosx
Answers
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
Thus the function can be rewritten as
f(x) = uv
Differentiating both sides n times with respect to x using Leibnitz theorem we get
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Answer:
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)}u
n
=cos(
2
nπ
+x)
\sf{v_1 = 3 {x}^{2} }v
1
=3x
2
\sf{v_2 = 6x }v
2
=6x
\sf{v_3 = 6 }v
3
=6
\sf{v_n= 0 \: \: \: for \: \: n > 3 }v
n
=0forn>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) }f
n
(x)
\sf{ = (uv)_n}=(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 + .. \: .. }=
n
C
0
u
n
v+
n
C
1
u
n−1
v
1
+
n
C
2
u
n−2
v
2
+
n
C
3
u
n
v
3
+
n
C
4
u
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 + .. \: ..=
n
C
0
cos(
2
nπ
+x)x
3
+
n
C
1
cos(
2
(n−1)π
+x)3x
2
+
n
C
2
cos(
2
(n−2)π
+x)6x+
n
C
3
cos(
2
(n−3)π
+x)6+
n
C
4
cos(
2
(n−4)π
+x).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=
n
C
0
cos(
2
nπ
+x)x
3
+
n
C
1
cos(
2
(n−1)π
+x)3x
2
+
n
C
2
cos(
2
(n−2)π
+x)6x+
n
C
3
cos(
2
(n−3)π
+x)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)=x
3
cos(
2
nπ
+x)+3nx
2
cos(
2
(n−1)π
+x)+3n(n−1)xcos(
2
(n−2)π
+x)+n(n−1)(n−2)cos(
2
(n−3)π
+x)
━━━━━━━━━━━━━━━━
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