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प्रश्न
If `sqrt(y + x) + sqrt(y - x)` = c, show that `"dy"/"dx" = y/x - sqrt(y^2/x^2 - 1)`.
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उत्तर
`sqrt(y + x) + sqrt(y - x)` = c
Differentiating both sides w.r.t. x, we get
`(1)/(2sqrt(y + x))."d"/"dx"(y + x) + (1)/(2sqrt(y - x))."d"/"dx"(y - x)` = 0
∴ `(1)/sqrt(y + x).(dy/dx + 1) + (1)/sqrt(y - x).(dy/dx - 1)` = 0
∴ `(1)/sqrt(y + x)."dy"/"dx" + (1)/sqrt(y + x) + (1)/sqrt(y - x)."dy"/"dx" - (1)/sqrt(y - x)` = 0
∴ `(1/sqrt(y + x) + 1/sqrt(y - x))"dy"/"dx" = (1)/sqrt(y - x) - 1/sqrt(y + x)`
∴ `[(sqrt(y - x) + sqrt(y + x))/(sqrt(y + x).sqrt(y - x))]"dy"/"dx" = (sqrt(y + x) + sqrt(y - x))/(sqrt(y - x).sqrt(y + x)`
∴ `"dy"/"dx" = (sqrt(y + x) + sqrt(y - x))/(sqrt(y + x).sqrt(y - x)`
= `= (sqrt(y + x) + sqrt(y - x))/(sqrt(y + x)+ sqrt(y - x)) xx (sqrt(y + x) + sqrt(y - x))/(sqrt(y + x) - sqrt(y - x)`
= `((sqrt(y + x) - sqrt(y - x)^2))/((y + x) - (y - x)`
= `(y + x + y - x - 2sqrt(y + x).sqrt(y - x))/(y + x - y + x)`
= `(2y - 2sqrt(y^2 - x^2))/(2x)`
= `(2y)/(2x) - (2sqrt(y^2 - x^2))/(2x)`
= `y/x - sqrt((y^2 - x^2)/x^2)`
∴ `"dy"/"dx" = y/x - sqrt(y^2/x^2 - 1)`
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