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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Dec 2007 00:38:35 IST
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Hello everyone! If y = (sinx)cosx + (cosx)sinx, prove that dy/dx = (sinx)cosx . [cotx cosx ? sinx (log sinx)] + (cosx)sinx. [cosx (log cosx) ? sinx tanx]
(CBSE 2004C) The answer that I got was dy/dx = [(sinx)cosx. (cosx)sinx] [cotx cosx ? sinx (logsinx) + cosx (log cosx) ? sinx tanx] Kindly explain why I am getting it different. I?ll be grateful to you. Thanks!
With regards
Ashish
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'Your eyes show the strength of your soul.' |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Dec 2007 00:40:23 IST
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That question mark (?) was actually meant to be the dot (.) for multiplication. I don't know what the problem is but every time it occurs. Any way out?
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firstly d correct ans is [sinx^cosx{cotxcosecx-sinxlogsinx} +cosx^sinx{cosxlogcosx+tanxsecx}] take p= sinx^cosx then differentiate it u will get d first expression and then q= cosx^sinx . and add the outcome.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Jan 2008 16:17:40 IST
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The Question is: If y = (sinx)cosx + (cosx)sinx, prove that dy/dx = (sinx)cox.[cotx.cosx - sinx(log sinx)] + (cosx)sinx.[cosx (log cosx) - sinx.tanx]
The answer is: Consider u = (sinx)cosx and v = (cosx)sinx Hence y = u + v Differentiating with respect to x, we get dy/dx = du/dx + dv/dx ........(1)
Now consider u = (sinx)cosx Taking log on both sides, we get log(u) = log(sinxcosx) log(u) = cosx.log(sinx) Differentiating with respect to x, we get 1/u . du/dx = -sinx.log(sinx) + (cosx/sinx).cosx du/dx = u.[cotx.cosx - sinx.log(sinx)] du/dx = (sinx)cosx[cotx.cosx - sinx.log(sinx)].......(2)
Now consider v = (cosx)sinx Taking log on both sides, we get log(v) = log(cosxsinx) log(u) = sinx.log(cosx) Differentiating with respect to x, we get 1/v . dv/dx = cosx.log(cosx) + (sinx/cosx)(-sinx) dv/dx = v.[cox.log(cosx) - tanx.sinx] dv/dx = (cosx)sinx[cox.log(cosx) - tanx.sinx]......(3)
Now from (1), we get dy/dx = (2) + (3) Hence dy/dx = (sinx)cosx[cotx.cosx - sinx.log(sinx)] + (cosx)sinx[cox.log(cosx) - tanx.sinx]
Hence proved.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Jan 2008 16:18:35 IST
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I hope u got it. If not please come forward and ask any doubt u hav!!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Jan 2008 17:19:30 IST
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Hello rvinitra!
Perfect answer! Alas! I have already accepted an answer. You deserve your first salute!
Thanks!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Jan 2008 21:47:23 IST
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thank u
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