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Find the coordinates of the point on the...

Find the coordinates of the point on the curve `y=x/(1+x^2)` where the tangent to the curve has the greatest slope.

Text Solution

Verified by Experts

The correct Answer is:
` x = 0, y = 0 `
Given, `" " y = (x)/(1 +x ^(2))`
`rArr " " (dy)/(dx) = (( 1 + x ^(2) ) * 1 - x ( 2x )) /(( 1 + x ^(2))) = (1- x ^(2))/((1 + x^(2))^(2))`
Let ` " " (dy)/(dx) = g (x)" "` [ i.e. slope of tangent ]
`therefore " "g(x) = (1 - x ^(2))/(( 1+ x ^(2))) `
`rArr g ' (x) = ((1 + x ^(2))^(2) * ( -2x ) - (1 - x ^(2)) * 2 ( 1+ x ^(2))* 2x )/( ( 1 + x ^(2)) ^(4))`
` " " = (-2x (1 + x ^(2)) [ (1 + x ^(2)) + 2 ( 1- x ^(2))]) /((1 + x ^(2)) ^(4)) = (- 2x (3- x ^(2)))/( (1 + x ^(2)) ^(3))`
For greater or least values of m , we should have
`g' (x) = 0 rArr x = 0 , x = pm sqrt 3 `
Now,
`g''(x) = ((1 + x ^(2))^(3) ( 6x ^(2) - 6) - ( 2 x ^(3) - 6x ) * 3 ( 1+ x ^(2)) ^(2 ) * 2x )/( (1 + x ^(2)) ^(6))`
At ` " "x = 0, g''(x) = - 6 lt 0`
` therefore g ' (x)` has a maximum value at ` x = 0`
`rArr (x = 0, y = 0 ) ` is the required point at which tangent to the curve has the greatest slope.
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