The equation of tangent to the curve `y=int_(x^2)^(x^3)(dt)/sqrt(1+t^2)` at `x=1` is `sqrt(3)x+1=y` (b) `sqrt(2)y+1=x` `sqrt(3)x+y=1` (d) `sqrt(2)y=x`

The equation of the tangent to the curve y= int_(x^4)^(x^6) (dt)/( sqrt( 1+t^2) ) at x=1 is

Find the equation of tangent to y=int_(x^(2))^(x^(3))(dt)/(sqrt(1+t^(2))) at x=1

equation of tangent to y=int_(x^(2)rarr x^(3))(dt)/(sqrt(1+t^(2))) at x=1 is:

Find the equation of tangent to the curve y=sin^(-1)(2x)/(1+x^(2)) at x=sqrt(3)

The slope of the tangent of the curve y=int_(x)^(x^(2))(cos^(-1)t^(2))dt at x=(1)/(sqrt2) is equal to

Equation of a line which is tangent to both the curve y=x^(2)+1 and y=x^(2) is y=sqrt(2)x+(1)/(2) (b) y=sqrt(2)x-(1)/(2)y=-sqrt(2)x+(1)/(2)(d)y=-sqrt(2)x-(1)/(2)

The slope of the tangent to the curve tan y = ((sqrt(1 + x)) - (sqrt(1 -x)))/(sqrt( 1+ x) + (sqrt (1 -x)) at x = 1/2 is

A circle C of radius 1 is inscribed in an equilateral triangle PQR . The points of contact of C with the sides PQ, QR, RP and D, E, F respectively. The line PQ is given by the equation sqrt(3) +y-6=0 and the point D is ((sqrt(3))/(2), 3/2) . Equation of the sides QR, RP are : (A) y=2/sqrt(3) x + 1, y = 2/sqrt(3) x -1 (B) y= 1/sqrt(3) x, y=0 (C) y= sqrt(3)/2 x + 1, y = sqrt(3)/2 x-1 (D) y=sqrt(3)x, y=0

Identify the curve sqrt((x+1)^(2)+y^(2))+ sqrt(x^(2)+(y-1)^(2))-2 =0