n- th derivative of sin 6x cos 4x
Answers
Step-by-step explanation:
SUCCESSIVE DIFFERENTIATION
Aditya S. asked • 06/30/18
Find the nth derivative of Cos^6x
Cos6x nth derivative
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1 Expert Answer
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Bobosharif S. answered • 06/30/18
TUTOR 4.4 (32)
Mathematics/Statistics Tutor
SEE TUTORS LIKE THIS
f(x)=cos6x
f'(x)=-6cos5(x) sin(x)=-3cos4(x)sin(2x)
f''(x)=-6(-5cos4(x)sin2(x)+cos6(x) )=6cos4(x)(2-3cos(2x))
f'''(x)=-120cos3(x) sin3(x)+60cos5(x) sin(x)+36cos5(x) sin(x)
=12cos3(x)(-1 + 9cos(2x))sin(x)
f(iv)(x)=6cos2(x)(11-22cos(2x)+27cos(4x))
f(v)(x)=-3(5sin(2x)+64sin(4x)+81sin(6x))
f(vi)(x)=-6(5cos(2x)+128cos(4x)+243cos(6x))
f(7)(x)=12(5sin(2x)+256sin(4x)+729sin(6x))
f(8)(x)=24(5cos(2x)+512cos(4x)+2187cos(6x) )
f(2n-1)(x)=(-1)n+13*2n-2(5sin(2x)+2n+3sin(4x)+3n+1sin(6x)), n=3,4,..
f(2n)(x)=(-1)n3*2n-1(5cos(2x)+2n+4cos(4x)+3ncos(6x)),n=3,4,...
Step-by-step explanation:
Here the given expression
\displaystyle \sf{ \sin 6x \cos 4x }sin6xcos4x
= \displaystyle \sf{ \frac{1}{2} \bigg( 2 \sin 6x \cos 4x\bigg) }=
2
1
(2sin6xcos4x)
= \displaystyle \sf{ \frac{1}{2} \bigg( \sin 10x + \sin 2x\bigg) }=
2
1
(sin10x+sin2x)
Now
\displaystyle \sf{ \frac{ {d}^{n} }{d {x}^{n} } \bigg( \sin 6x \cos 4x \bigg)}
dx
n
d
n
(sin6xcos4x)
= \displaystyle \sf{ \frac{1}{2} \times \frac{ {d}^{n} }{d {x}^{n} }( \sin 10x) + \frac{1}{2} \times \frac{ {d}^{n} }{d {x}^{n} }( \sin 2x)}=
2
1
×
dx
n
d
n
(sin10x)+
2
1
×
dx
n
d
n
(sin2x)
\displaystyle \sf{= \frac{1}{2} \times {10}^{n} \sin \bigg( \frac{n\pi}{2} + 10x\bigg) + \frac{1}{2} \times {10}^{n} \sin \bigg( \frac{n\pi}{2} + 2x\bigg) }=
2
1
×10
n
sin(
2
nπ
+10x)+
2
1
×10
n
sin(
2
nπ
+2x)
\displaystyle \sf{= \frac{ {10}^{n}}{2} \times \Bigg[ \sin \bigg( \frac{n\pi}{2} + 10x\bigg) + \sin \bigg( \frac{n\pi}{2} + 2x\bigg)\Bigg] }=
2
10
n
×[sin(
2
nπ
+10x)+sin(