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

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

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 the curve y=sin^(-1)(2x)/(1+x^(2)) at x=sqrt(3)

int_(0)^(1)t^(5)*sqrt(1-t^(2))*dt

Find the deriative of the function f (x) = int _(0) ^(x) sqrt (1+ t ^(2)) dt.

int_(0)^(1)t^(2)sqrt(1-t)*dt