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प्रश्न
If y = log (log 2x), show that xy2 + y1 (1 + xy1) = 0.
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उत्तर
y = log (log 2x)
∴ `"dy"/"dx" = "d"/"dx"[log(log2x)]`
= `(1)/"log2x"."d"/"dx"(log2x)`
= `(1)/"log2x" xx (1)/(2x)."d"/"dx"(2x)`
= `(1)/"log2x" xx (1)/(2x) xx 2`
∴ `"dy"/"dx" = (1)/(xlog2x)`
∴ `(log2x)."dy"/"dx" = (1)/x` ...(1)
Differentiating both sides w.r.t. x, we get
`(log2x)."d"/"dx"(dx/dy) + "dy"/"dx"."d"/"dx"(log2x) = "d"/"dx"(1/x)`
∴ `(log2x).(d^2y)/(dx^2) + "dy"/"dx".(1)/(2x)."d"/"dx"(2x) = -(1)/x^2`
∴ `(log2x).(d^2y)/(dx^2) + "dy"/"dx".(1)/(2x) xx 2 = -(1)/x^2`
∴ `(log2x).(d^2y)/(dx^2) + (1)/x."dy"/"dx" = (1)/x.(1)/x`
∴ `(log2x).(d^2y)/(dx^2) + [(log2x)."dy"/"dx"]"dy"/"dx" = -(1)/x[(log2x)."dy"/"dx"]` ...[By (1)]
Dividing throughout by log 2x, we get
`(d^2y)/(dx^2) + (dy/dx)^2 = -(1)/x"dy"/"dx"`
∴ `x(d^2y)/(dx^2) + x(dy/dx)^2 = -"dy"/"dx"`
∴ `x(d^2y)/(dx^2) + "dy"/"dx" + x(dy/dx)^2` = 0
∴ `x(d^2y)/(dx^2) + "dy"/"dx" (1 + xdy/dx)` = 0
∴ xy2 + y1 (1 + xy1) = 0.
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