Question

Find nth derivative of sinx cos3x

Answer 1

Sinx.Cos3x
Sinx (4Cos^3x - 3Cosx)
Sinx Cosx (4Cos^2x - 3)

Answer 2

The nth derivative of sin x cos 3x

$\displaystyle \sf{= \frac{1}{2} \bigg[ {4}^{n} \sin \bigg( \frac{n\pi}{2} + 4x\bigg) - {2}^{n} \sin \bigg( \frac{n\pi}{2} + 2x\bigg) \bigg] }$

Given :

sin x cos 3x

To find :

The nth derivative of sin x cos 3x

Formula :

The nth order derivative of sin ax

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

Solution :

Step 1 of 3 :

Write down the given expression

The given expression is sin x cos 3x

Step 2 of 3 :

Express as sum of two Trigonometric function

$\displaystyle \sf{ \sin x \cos 3x }$

$= \displaystyle \sf{ \frac{1}{2} \bigg( 2 \sin x \cos 3x\bigg) }$

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

Step 3 of 3 :

Find nth derivative of sin x cos 3x

$\displaystyle \sf{ \frac{ {d}^{n} }{d {x}^{n} } \bigg( \sin x \cos 3x \bigg)}$

$= \displaystyle \sf{ \frac{1}{2} \times \frac{ {d}^{n} }{d {x}^{n} }( \sin 4x) - \frac{1}{2} \times \frac{ {d}^{n} }{d {x}^{n} }( \sin 2x)}$

$\displaystyle \sf{= \frac{1}{2} \times {4}^{n} \sin \bigg( \frac{n\pi}{2} + 4x\bigg) - \frac{1}{2} \times {2}^{n} \sin \bigg( \frac{n\pi}{2} + 2x\bigg) }$

$\displaystyle \sf{= \frac{1}{2} \bigg[ {4}^{n} \sin \bigg( \frac{n\pi}{2} + 4x\bigg) - {2}^{n} \sin \bigg( \frac{n\pi}{2} + 2x\bigg) \bigg] }$

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