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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Nov 2006 18:34:05 IST
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hi
this is the solution 4 question 3rd
y=f(3x+4/5x+6) nd f'(x)=tanx2
dy/dx=f'(3x+4/5x+6) d/dx(3x+4/5x+6)
dy/dx=tan(3x+4/5x+6)^2 * (-2/5x+6)^2
thus dy/dx= {-2/(5x+6)^2 }* {tan(3x+4/5x+6)^2}
hope my answer is correct
ruhi
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this reply: 5 points
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