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Find the point on the curve `y=x/(1+x^2)`, where the tangent to the curve has the greatest slope. |
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Answer» slope=`dy/dx=d/dx(x/(1+x^2))` `=((1+x^2)*1-x(2x))/(1+x^2)^2` `=(1-x^2)/(1+x^2)^2` `dy/dx` is maximum when`(x/(1+x^2))^2` is minimum `x=0,y=0` |
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