dy/dx

Asked by  | 12th Jun, 2013, 10:13: PM

Expert Answer:

y= logx+(logx+((logx...infinty )power 1/2)power 1/2)power1/2
 
y = logx + y^(1/2)
Now, differentiating wrt x
dy/dx = 1/x +1/( 2y^(1/2))
dy/dx= 1/x + 1/2 [logx+(logx+((logx...infinty )power 1/2)power 1/2)power1/2]^1/2]

Answered by  | 13th Jun, 2013, 04:10: AM

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