Calculus - Differentiation

13

Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-03-07 00:01:17

y=âˆš(cosx+y)
y²-y=cos x
2y-1=-sin xdx/dy
(2y-1)dy/dx+sinx=0

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MATRIX ·2009-03-07 00:02:02

i want to knw is there any other method than this method....

y√cosx+y

y²=cosx+y

Differentiating.........

2y.dy/x=-sinx+dy/dx

therefore.....

(2y-1)dy/dx+sinx=0.....

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13

Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-03-07 00:02:21

[3]

bhaaiya...u type so fast....!!

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62

Lokesh Verma ·2009-03-07 00:05:41

haha abhirup :)

I will delete mine ;)

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Lokesh Verma ·2009-03-07 00:06:21

yes prajith

this can be done b ychain rule alone

think for a moment :)

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MATRIX ·2009-03-07 00:07:53

bhayya actually i knw the method of y²=cosx+y.....i have typed it without seeing abhirups answer.........

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Lokesh Verma ·2009-03-07 00:17:31

y=\sqrt{cosx+\sqrt{cosx+\sqrt{cosx......}}} \Rightarrow \frac{dy}{dx} =\frac{1}{2 \sqrt{cosx+\sqrt{cosx+\sqrt{cosx......}}}}\times \frac{d}{dx}\left\{{cosx+\sqrt{cosx+\sqrt{cosx......}}} \right\} \Rightarrow \frac{dy}{dx} =\frac{1}{2 \sqrt{cosx+\sqrt{cosx+\sqrt{cosx......}}}}\times\left\{-sinx + \frac{dy}{dx} \right\}thus, \frac{dy}{dx}=\frac{1}{2y}\times\left\{-\sin x +\frac{dy}{dx} \right\}