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Fine the point on the curve y=x/(1+x^{2) } where the tangent to the curve has the greatest slope

y = x / 1 + x

^{2 }

Slope of the tangent to this curve at any point is given by

M = dy/dx = ( 1 + x

^{2}) – 2x(x)/( 1 + x

^{2})

^{2}(using quotient rule)

= 1- x

^{2}/ (1 + x

^{2})

^{2}

This is what we need to maximize.

Let’s differentiate M w.r.t x using quotient rule.

dM/dx = -2x( 1 + x

^{2})

^{2}- 2( 1 + x

^{2}) 2x ( 1 – x

^{2}) / ( 1 + x

^{2})

^{4}

= -2x( 1 + x

^{2}) [ 1 + x

^{2}+ 2 ( 1 – x

^{2}) ] / ( 1 + x

^{2})

^{4}

= -2x( 1 + x

^{2})( 3 – x

^{2}) / ( 1 + x

^{2})

^{4 }

At maxima or minima , dM/dx = 0

So we get -2x( 1 + x

^{2})( 3 – x

^{2}) = 0

Which gives x = 0 , √3 , - √3

On doing the first derivative test , it can be seen that as the function crosses these turning points , at x = 0 , its value changes from +ve to –ve and at other two points , the value change from –ve to +ve . So 0 is the point of maxima and others are the point of minima .

So the x coordinate of the point at which curve has maximum slope = 0 .

Substituting x = 0 in the equation of curve , we get y = 0.

So the required point is ( 0 , 0 ) .

Hope that helps.