Among the following points, at which point dydx of y=x1+x2 will be maximum

y=x1+x2 m=dydx=(1+x2)−x(2x)(1+x2)=1−x2(1+x2)2 dmdn=(1+x2)2(−2x)−(1−x2)2(1+x2)(2x)(1+x2)4 =2x(1+x2)[−1−x2−2+2x2](1+x2)4 =2x(x2−3)(x2+1)3 dydx=m, m is maximum at x=0 (x,y)=(0,0)

Among the following points, at which point dydx of y=x1+x2 will be maximum

y=x1+x2 m=dydx=(1+x2)−x(2x)(1+x2)=1−x2(1+x2)2 dmdn=(1+x2)2(−2x)−(1−x2)2(1+x2)(2x)(1+x2)4 =2x(1+x2)[−1−x2−2+2x2](1+x2)4 =2x(x2−3)(x2+1)3 dydx=m, m is maximum at x=0 (x,y)=(0,0)

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Among the following points, at which point $\frac{d y}{d x}$ of $y = \frac{x}{1 + x^{2}}$ will be maximum

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$(0, 0)$

$(1, \frac{1}{2})$

$(- 1, \frac{- 1}{2})$

$(2, \frac{2}{5})$

answer is D.

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Detailed Solution

$y = \frac{x}{1 + x^{2}} m = \frac{d y}{d x} = \frac{(1 + x^{2}) - x (2 x)}{(1 + x^{2})} = \frac{1 - x^{2}}{{(1 + x^{2})}^{2}} \frac{d m}{d n} = \frac{{(1 + x^{2})}^{2} (- 2 x) - (1 - x^{2}) 2 (1 + x^{2}) (2 x)}{{(1 + x^{2})}^{4}} = \frac{2 x (1 + x^{2}) [- 1 - x^{2} - 2 + 2 x^{2}]}{{(1 + x^{2})}^{4}} = \frac{2 x (x^{2} - 3)}{{(x^{2} + 1)}^{3}}$