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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Show that dydxdydx=yx in the following, where a and p are constant: xpy4 = (x + y)p+4, p ∈ N

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प्रश्न

Show that `bb("dy"/"dx" = y/x)` in the following, where a and p are constant:

xpy4 = (x + y)p+4, p ∈ N

बेरीज
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उत्तर

xpy4 = (x + y)p+4 

Taking log

log(xpy4) = log(x + y)p+4

logxp + logy4 = (p + 4) log(x + y)

p log x + 4 log y = (p + 4) log(x + y)

Differentiating both sides w.r.t. x, we get

`p."d"/"dx"logx + 4*"d"/"dx"logy = (p + 4)"d"/"dx"log(x + y)`

`p/x + 4(1)/y"dy"/"dx" = (p + 4)(1)/(x + y)(1 + "dy"/"dx")`

`"p"/x + 4/y"dy"/"dx" = ((p + 4))/((x + y)) + (p + 4)/((x + y))"dy"/"dx"`

`"dy"/"dx"[4/y - ((p + 4))/((x + y))] = (p + 4)/(x + y) - p/x`

`"dy"/"dx"[(4(x + y) -y(p + 4))/(y(x + y))] = (x(p + 4) -p(x + y))/(x(x + y)`

`"dy"/"dx"[(4x + 4y - py - 4y)/(y(x + y))] = (px + 4x - px - py)/(x(x + y)`

`"dy"/"dx"[(4x - py)/y] = (4x - py)/x`

`"dy"/"dx" = y/x`

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पाठ 1: Differentiation - Exercise 1.3 [पृष्ठ ४०]

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