IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

ganesha1991

(cos4a)/x + (sin4a)/y=1/(x+y)

y(cos⁴ a) + x (sin⁴ a) = xy /x+y
differentiating
we get
dy/dx cos⁴ a + sin⁴ a = [y+xdy/dx ] [x+y] - [1+dy/dx] [ xy] /(x+y)²
dy/dxcos⁴ a+sin⁴ a = [xy + x² dy/dx + y² +xydy/dx - xy - xydy/dx ] /(x+y)²
dy/dx cos⁴a + sin⁴ a = [x² dy/dx + y² ] /(x+y)²
dy/dxcos⁴ a (x+y)² +sin⁴ a(x+y)² = x² dy/dx + y²
dy/dx [ cos⁴ a(x+y)² -x² ] = y² - sin⁴ a (x+y)²
dy/dx = [y² -sin⁴ a(x+y)² ] / [cos⁴ a(x+y)² -x² ]

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