AP ICET 2017 Shift 1 Paper

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Question : 135 of 200

Marks: +1, -0

y=\cot^{-1}(\;\;{1-3x^{2}}/{3x-x^{3}})⇒\;\;{dy}/{dx}=

$-3/{1+x^2}$
$3/{1+x^2}$
$-3/{1-x^2}$
$-3/{1-x^2}$

Solution:

We know that

d/{dx}=\cot^{-1}x(\1/{1+x^2})

y=\cot^{-1}({1-3x^{2}}/{3x-x^{3}})

Let

z={1-3x^2}/{3x-x^3}

,then

y=\cot^{-1}z

{dz}/{dx}={(3x-x^{3})(-6x)-(1-3x^{2})(3-3x^{2})}/{(3x-x^{3})^{2}}

={-18x^{2}+6x^{4}-3+12x^{2}-9x^{4}}/{(3x-x^{3})^{2}}

={-3x^{4}-6x^{2}-3}/{(3x-x^{3})^{2}}={-3(x^{4}+2x^{2}+1)}/{(3x-x^{3})^{2}}={-3(x^{2}+1)^{2}}/{(3x-x^{3})^{2}}

1+z^{2}=1+({1-3x^{2}}/{3x-x^{3}})^{2}={(3x-x^{3})^{2}+(1-3x^{2})^{2}}/{(3x-x^{3})^{2}}

={9x^{2}+x^{6}-6x^{4}+1+9x^{4}-6x^{2}}/{(3x-x^{3})^{2}}

={x^{6}+3x^{4}+3x^{2}+1}/{(3x-x^{3})^{2}}={(x^{2}+1)^{3}}/{(3x-x^{3})^{2}}

y=\cot^{-1}z

{dy}/{dx}=-{1}/{1+z^{2}}⋅{dz}/{dx}

=-{(3x-x^{3})^{2}}/{(x^{2}+1)^{3}}⋅{-3(x^{2}+1)^{2}}/{(3x-x^{3})^{2}}=\;\;{3}/{x^{2}+1}

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