d^20/dx^20(2cosx cos3x)= A. 2^20(cos2x – 2^20 cos 4x) B. 2^20(cos2x + 2^20 cos 4x) C. 2^20(sin2x – 2^20 sin 4x) D. 2^20(sin2x – 2^20 sin 4x)

Question

\(\frac{d^{20}}{dx^{20}}(2\,cosx\,cos\,3x)\)=

A. 2²⁰(cos2x – 2²⁰ cos 4x) 

B. 2²⁰(cos2x + 2²⁰ cos 4x) 

C. 2²⁰(sin2x – 2²⁰ sin 4x) 

D. 2²⁰(sin2x – 2²⁰ sin 4x)

Answer 1

Correct Answer is (B) (-2)²⁰ (cos 2x+2²⁰ cos 4x )

Given:

Let y=2 cos x cos 3x

\(2\,cos\,A\,cos\,B\)\(=cos(\frac{A+B}{2})+cos(\frac{A-B}{2})\)

So y=cos 2x+cos 4x

\(\frac{dy}{dx}=-2sin\,2x-4sin\,4x\)

= (-2)¹ (sin 2x+2¹ sin 4x )

\(\frac{d^2y}{dx^2}=-4cos\,2x-16cos\,4x\)

= (-2)² (cos 2x+2² cos 4x )

\(\frac{d^3y}{dx^3}=8cos\,2x+64cos\,4x\)

= (-2)³ (cos 2x+2³ cos 4x )

\(\frac{d^4y}{dx^4}=16cos\,2x+256cos\,4x\)

= (-2)⁴ (cos 2x+2⁴ cos 4x )

For every odd degree; differential = =(-2)ⁿ (cos 2x+2ⁿ cos 4x );n={1,3,5…}

For every even degree; differential =(-2)ⁿ (cos 2x+2ⁿ cos 4x );n={0,2,4…}

So,\(\frac{d^{20}y}{dx^{20}}\)\(=(-2)^{20}(cos2x+2^{20}cos4x)\)

= (-2)²⁰ (cos 2x+2²⁰ cos 4x );