If the slope of normal to the curve at a point is 5 , then at that point slope of tangent to the curve is

( b ) 5
( c )` 1/5`
( d ) `-1/5`

If the slope of the curve y=(ax)/(b-x) at the point (1, 1) is 2, then

Write the slope of the normal to the curve y=1/x at the point (3,1/3)

The slope of the tangent to the curve y = 3x^(2) - 5x + 6 at (1, 4) is

Find the slope of tangent of the curve y =4x at the point (-1,4).

The slope of the tangent to the curve at a point (x,y) on it is proportional to (x-2) if the slope of tangent to the curve at (10,-9) on it is -3 then the equation of the curve is :

The slope of the tangent line to the curve x+y=xy at the point (2,2) is

Find the slope of the tangent to the curve y(x^2+1)=x at the point (1, 1/2)

Find the slope of tangent of the curve y=x^(2) at point ((1)/(3),(1)/(9))

Find the slope of the tangent to the curve y=(x-3)/(x-5), x ne 5" at " x=10 .