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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