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Question Number 8996 by Basant007 last updated on 11/Nov/16

Find the nth derivative of sin^2 2x

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{nth}\:\mathrm{derivative}\:\mathrm{of}\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x} \\ $$

Commented by FilupSmith last updated on 13/Nov/16

y=sin^2 (2x) u=sin(2x) ⇒ du=2cos(2x)dx dx=(1/(2cos(2x)))du (dy/dx)=(d/dx)(u^2 ) (dy/dx)=(d/((1/(2cos(2x)))du))(u^2 ) (dy/dx)=2cos(2x)(d/du)(u^2 ) (dy/dx)=4cos(2x)u (dy/dx)=4cos(2x)sin(2x) cos(x)sin(x)=(1/2)sin(2x) (dy/dx)=2sin(4x) (d^2 y/dx^2 )=8cos(4x) (d^3 y/dx^3 )=−32sin(4x) (d^4 y/dx^4 )=−128cos(4x) I′m not sure if there is a general form for (d^n y/dx^n )

$${y}=\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right) \\ $$$${u}=\mathrm{sin}\left(\mathrm{2}{x}\right)\:\Rightarrow\:{du}=\mathrm{2cos}\left(\mathrm{2}{x}\right){dx} \\ $$$${dx}=\frac{\mathrm{1}}{\mathrm{2cos}\left(\mathrm{2}{x}\right)}{du} \\ $$$$\frac{{dy}}{{dx}}=\frac{{d}}{{dx}}\left({u}^{\mathrm{2}} \right) \\ $$$$\frac{{dy}}{{dx}}=\frac{{d}}{\frac{\mathrm{1}}{\mathrm{2cos}\left(\mathrm{2}{x}\right)}{du}}\left({u}^{\mathrm{2}} \right) \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2cos}\left(\mathrm{2}{x}\right)\frac{{d}}{{du}}\left({u}^{\mathrm{2}} \right) \\ $$$$\frac{{dy}}{{dx}}=\mathrm{4cos}\left(\mathrm{2}{x}\right){u} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{4cos}\left(\mathrm{2}{x}\right)\mathrm{sin}\left(\mathrm{2}{x}\right) \\ $$$$\mathrm{cos}\left({x}\right)\mathrm{sin}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{2}{x}\right) \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2sin}\left(\mathrm{4}{x}\right) \\ $$$$\: \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{8cos}\left(\mathrm{4}{x}\right) \\ $$$$\: \\ $$$$\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }=−\mathrm{32sin}\left(\mathrm{4}{x}\right) \\ $$$$\: \\ $$$$\frac{{d}^{\mathrm{4}} {y}}{{dx}^{\mathrm{4}} }=−\mathrm{128cos}\left(\mathrm{4}{x}\right) \\ $$$$\: \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{general}\:\mathrm{form} \\ $$$$\mathrm{for}\:\:\frac{{d}^{{n}} {y}}{{dx}^{{n}} } \\ $$

Answered by 123456 last updated on 14/Nov/16

y=sin^2 2x (dy/dx)=2sin 4x (d^2 y/dx^2 )=8cos 4x=8sin(4x+(π/2)) (d^3 y/dx^3 )=−32sin 4x=−32sin(4x+π) (d^4 y/dx^4 )=−128cos 4x=−128sin(4x+((3π)/2)) (d^5 y/dx^5 )=512sin 4x −−−−−−− 1→2=2∙4^0 ;0 2→8=2∙4^1 =2^3 ;(π/2) 3→32=2∙4^2 =2^5 ;π 4→128=2∙4^3 =2^7 ;((3π)/2) n→2∙4^(n−1) =2∙2^(2n−2) =2^(2n−1) ;(((n−1)π)/2) (d^n y/dx^n )= { ((sin^2 2x),(n=0)),((2^(2n−1) sin (4x+(((n−1)π)/2))),(n>0)) :}

$${y}=\mathrm{sin}^{\mathrm{2}} \mathrm{2}{x} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2sin}\:\mathrm{4}{x} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{8cos}\:\mathrm{4}{x}=\mathrm{8sin}\left(\mathrm{4}{x}+\frac{\pi}{\mathrm{2}}\right) \\ $$$$\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }=−\mathrm{32sin}\:\mathrm{4}{x}=−\mathrm{32sin}\left(\mathrm{4}{x}+\pi\right) \\ $$$$\frac{{d}^{\mathrm{4}} {y}}{{dx}^{\mathrm{4}} }=−\mathrm{128cos}\:\mathrm{4}{x}=−\mathrm{128sin}\left(\mathrm{4}{x}+\frac{\mathrm{3}\pi}{\mathrm{2}}\right) \\ $$$$\frac{{d}^{\mathrm{5}} {y}}{{dx}^{\mathrm{5}} }=\mathrm{512sin}\:\mathrm{4}{x} \\ $$$$−−−−−−− \\ $$$$\mathrm{1}\rightarrow\mathrm{2}=\mathrm{2}\centerdot\mathrm{4}^{\mathrm{0}} ;\mathrm{0} \\ $$$$\mathrm{2}\rightarrow\mathrm{8}=\mathrm{2}\centerdot\mathrm{4}^{\mathrm{1}} =\mathrm{2}^{\mathrm{3}} ;\frac{\pi}{\mathrm{2}}\: \\ $$$$\mathrm{3}\rightarrow\mathrm{32}=\mathrm{2}\centerdot\mathrm{4}^{\mathrm{2}} =\mathrm{2}^{\mathrm{5}} ;\pi \\ $$$$\mathrm{4}\rightarrow\mathrm{128}=\mathrm{2}\centerdot\mathrm{4}^{\mathrm{3}} =\mathrm{2}^{\mathrm{7}} ;\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$${n}\rightarrow\mathrm{2}\centerdot\mathrm{4}^{{n}−\mathrm{1}} =\mathrm{2}\centerdot\mathrm{2}^{\mathrm{2}{n}−\mathrm{2}} =\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} ;\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}} \\ $$$$\frac{{d}^{{n}} {y}}{{dx}^{{n}} }=\begin{cases}{\mathrm{sin}^{\mathrm{2}} \mathrm{2}{x}}&{{n}=\mathrm{0}}\\{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \mathrm{sin}\:\left(\mathrm{4}{x}+\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}}\right)}&{{n}>\mathrm{0}}\end{cases} \\ $$

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