Question

Find the nth derivative of sin 6x cos 4x.

Answer 1

SOLUTION

TO DETERMINE

The nth derivative of sin 6x cos 4x

FORMULA TO BE IMPLEMENTED

The nth order derivative of sin ax

$\displaystyle \sf{ \frac{ {d}^{n} }{d {x}^{n} }( \sin ax) = {a}^{n} \sin \bigg( \frac{n\pi}{2} + ax\bigg)}$

EVALUATION

Here the given expression

$\displaystyle \sf{ \sin 6x \cos 4x }$

$= \displaystyle \sf{ \frac{1}{2} \bigg( 2 \sin 6x \cos 4x\bigg) }$

$= \displaystyle \sf{ \frac{1}{2} \bigg( \sin 10x + \sin 2x\bigg) }$

Now

$\displaystyle \sf{ \frac{ {d}^{n} }{d {x}^{n} } \bigg( \sin 6x \cos 4x \bigg)}$

$= \displaystyle \sf{ \frac{1}{2} \times \frac{ {d}^{n} }{d {x}^{n} }( \sin 10x) + \frac{1}{2} \times \frac{ {d}^{n} }{d {x}^{n} }( \sin 2x)}$

$\displaystyle \sf{= \frac{1}{2} \times {10}^{n} \sin \bigg( \frac{n\pi}{2} + 10x\bigg) + \frac{1}{2} \times {10}^{n} \sin \bigg( \frac{n\pi}{2} + 2x\bigg) }$

$\displaystyle \sf{= \frac{ {10}^{n}}{2} \times \Bigg[ \sin \bigg( \frac{n\pi}{2} + 10x\bigg) + \sin \bigg( \frac{n\pi}{2} + 2x\bigg)\Bigg] }$

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Answer 2

Answer:

SOLUTION

TO DETERMINE

The nth derivative of sin 6x cos 4x

FORMULA TO BE IMPLEMENTED

The nth order derivative of sin ax

\displaystyle \sf{ \frac{ {d}^{n} }{d {x}^{n} }( \sin ax) = {a}^{n} \sin \bigg( \frac{n\pi}{2} + ax\bigg)}

dx

n

d

n

(sinax)=a

n

sin(

2

nπ

+ax)

EVALUATION

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