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Question
If ey (x + 1) = 1, show that `(d^2y)/(dx^2) = (dy/dx)^2`.
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Solution
Given, ey (x + 1) = 1 ....(1)
Differentiating (1) w.r.t. x, we get
`d/dx e^y (x + 1) = d/dx (1)`
`e^y d/dx (x + 1) + (x + 1) d/dx e^y = 0`
`e^y(1) + (x + 1) e^y dy/dx = 0`
`e^y + (x + 1) e^y dy/dx = 0`
`(x + 1)e^y dy/dx = -e^y`
`(x + 1) dy/dx = -1`
`dy/dx = (-1)/(x + 1)`
Now square both sides:
`(dy/dx)^2 = 1/(x + 1)^2`
Differentiating both sides again with respect to x,
`(d^2 y)/dx^2 = - d/dx 1/(x + 1)`
= `- d/dx (x + 1)^-1`
= `-(-1) (x + 1)^(-1-1)`
= 1 (x + 1)−2
`(d^2 y)/dx^2 = 1/(x + 1)^2`
Hence, `(d^2 y)/dx^2 = (dy/dx)^2`
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