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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

If y = log (log 2x), show that xy2 + y1 (1 + xy1) = 0.

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प्रश्न

If y = log (log 2x), show that xy2 + y1 (1 + xy1) = 0.

बेरीज
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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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पाठ 1: Differentiation - Exercise 1.5 [पृष्ठ ६०]

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