If e^y(x+1)=1, show that d^2/dx^2=(dy/dx)^2
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Given,
we have to proof :
e^y (x + 1) = 1
=> e^y = 1/(x + 1)
take log both sides,
=> y = log{1/(x + 1)}
differentiate y with respect to x,
=> dy/dx = d[log{1/(x + 1)}]/dx
=> dy/dx = 1/{1/(x + 1)} × -1/(x +1)²
=> dy/dx = (x + 1)/1 × -1/(x +1)²
=> dy/dx = -1/(x +1) ------(1)
squaring both sides,
(dy/dx)² = 1/(x + 1)² --------(2)
from equation (1),
dy/dx = -1/(x +1) =
differentiate once again,
d²y/dx² =
=> d²y/dx² =
e.g., d²y/dx² = 1/(x + 1)² -----(3)
from equations (2) and (3),
d²y/dx² = (dy/dx)²
we have to proof :
e^y (x + 1) = 1
=> e^y = 1/(x + 1)
take log both sides,
=> y = log{1/(x + 1)}
differentiate y with respect to x,
=> dy/dx = d[log{1/(x + 1)}]/dx
=> dy/dx = 1/{1/(x + 1)} × -1/(x +1)²
=> dy/dx = (x + 1)/1 × -1/(x +1)²
=> dy/dx = -1/(x +1) ------(1)
squaring both sides,
(dy/dx)² = 1/(x + 1)² --------(2)
from equation (1),
dy/dx = -1/(x +1) =
differentiate once again,
d²y/dx² =
=> d²y/dx² =
e.g., d²y/dx² = 1/(x + 1)² -----(3)
from equations (2) and (3),
d²y/dx² = (dy/dx)²
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