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Differential Calculus

Scorching goIITian

 Joined: 8 Feb 2007 Post: 217
29 May 2007 10:34:55 IST
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4
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Differentiate-
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Differential Calculus

Find dy/dx-

tanx.tan2x.tan3x

Blazing goIITian

Joined: 13 Jan 2007
Posts: 1289
29 May 2007 10:38:22 IST
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just apply product rule temr by term

Blazing goIITian

Joined: 26 Oct 2006
Posts: 530
29 May 2007 10:39:49 IST
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tan(x+2x)=(tanx + tan2x)/(1-tanx tan2x)
solving it..we get....................
tanx.tan2x.tan3x=tan3x - tanx - tan2x
soo nw....
differentiation of (tan3x - tanx - tan2x)
= 3sec^2 3x - sec^2 x - 2sec^2 2x

Forum Expert
Joined: 19 Oct 2006
Posts: 1966
29 May 2007 13:43:08 IST
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hii

well ruhi has shown a good trick .. though she did some mistake ..

so going with the trick

tan(x+2x) = (tanx + tan2x)/(1-tanx tan2x)
solving it, we get
tanx.tan2x.tan3x=tan3x - tanx - tan2x

and thus dy/dx = 3.sec23x - sec2x - 2.sec22x

cheers

Blazing goIITian

Joined: 12 May 2007
Posts: 328
29 May 2007 14:21:05 IST
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another easy way is applying log on both sides

rate if useful

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