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if [tex]y= \sqrt{sin (x)+\sqrt{sin (x)+\sqrt{sin (x)+....} } } [/tex]
till infinity
then [tex] \frac{dy}{dx}=? [/tex]

Answer :

manitkapoor2
[tex]y= \sqrt{sin x + \sqrt{sin x + \sqrt{sin x + ..} } } [/tex]
[tex]y^2-sinx=\sqrt{sin x +\sqrt{sin x +\sqrt{sin x + } .. } }=y [/tex]
diff. w.r.t x
[tex]2y \frac{dy}{dx}-cosx= \frac{dy}{dx} [/tex]
[tex] \frac{dy}{dx}= \frac{cos x}{2y +1} [/tex]
kopal
y = undr root= ur = ur of sinx +ur of sinx +ur of sinx.+............
y*y - sinx= "    "    "    "'    "     "
2y = dy/dx -cosx = dy/dx
dy/dx = cosx / 2y+1.
hope this makes u understand.!!!!!

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