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प्रश्न
If y `sqrt(x^2 + 1) = log sqrt(x^2 + 1) - x`, show that `(x^2 + 1)(dy)/(dx) + xy + 1 = 0.`
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उत्तर
Given:
y `sqrt(x^2 + 1) = log (sqrt(x^2 + 1) - x)`
Differentiate the Left-Hand Side:
Using the product rule (uv)′ = u′v + uv′:
Let u = y and v = `sqrt(x^2 + 1)`
`d/dx (y sqrt(x^2 + 1)) = (dy)/(dx) . sqrt(x^2 + 1) + y . d/dx (sqrt(x^2 + 1))`
= `sqrt(x^2 + 1) (dy)/(dx) + y . (1/(2sqrt(x^2 + 1)) . 2x)`
= `sqrt(x^2 + 1) (dy)/(dx) + (xy)/sqrt(x^2 + 1)` ...(i)
Differentiate the Right-Hand Side:
Using the chain rule for log(u):
`d/dx [log (sqrt(x^2 + 1) - x)] = 1/(sqrt(x^2 + 1) - x) . d/dx (sqrt(x^2 + 1) - x)`
= `1/(sqrt(x^2 + 1) - x) . (x/sqrt(x^2 + 1) - 1)`
Take the LCM in the bracket:
= `1/(sqrt(x^2 + 1) - x) . ((x - sqrt(x^2 + 1))/sqrt(x^2 + 1))`
= `1/(sqrt(x^2 + 1) - x) . ((-sqrt(x^2 + 1) - x)/sqrt(x^2 + 1))`
= `-1/(sqrt(x^2 + 1)` ...(ii)
Equate LHS and RHS
`sqrt(x^2 + 1) (dy)/(dx) + (xy)/sqrt(x^2 + 1) = -1/(sqrt(x^2 + 1)`
Multiply the entire equation by `sqrt(x^2 + 1)` to clear the denominators:
`(sqrt(x^2 + 1) . sqrt(x^2 + 1)) (dy)/(dx) + xy = -1`
`(x^2 + 1) (dy)/(dx) + xy = -1`
`(x^2 + 1) (dy)/(dx) + xy + 1 = 0`
Hence proved
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