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Question

Find ddx(sinmx.cosnx)
 


Solution

Let y=sinmx.cosnx dydx=ddx[(sinx)m.(cosx)n]=(sinx)m.ddx(cosx)n+(cosx)n.ddx(sinx)m=(sinx)m.n(cosx)n1.ddxcosx+      (cosx)nm(sinx)m1.ddxsin x=(sinx)m.n(cosx)n1x.(sinx)+mcosn.m(sinx)m1cosx=nsinmx.cosn1x.(sinx)+mcosnx.sinm1x.cosx=n.sinmx.sinx.cosnx.1cosx+m.sinmx1sinx.cosnx.cosx=n.sinmx.cosnx.tanx+msinmx.cosnx.cotx=sinmx.cosnx[mcotxntanx]

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