If `f(y)=e^y,g(y)=y,y>0, and F(t)=int_0^t f(t-y)g(y) dy`, then

Iff(y)=e^(y),g(y)>0, and F(t)=int_(0)^(t)t(t-y)dt, then F(t)=e^(t)-(1+t)F(t)=te^(t)F(t)=te^(-1)(d)F(t)=1-e^(t)(1+t)

If int_(y)^(y)cos t^(2)dt=int_(0)^(x^(2))(sin t)/(t)dx, the prove that (dy)/(dx)=(2sin x^(2))/(x cos y^(2))

If x = f(t) , y = g(t) possess second order derivatives for all t in (a, b) and f'(t) is invertible [f'(t)ne0] , then prove that : (d^(2)y)/(dx^(2))=(f'(t)g''(t)-g'(t)f''(t))/([f'(t)]^(3)) .

If f((x)/(y))=(f(x))/(f(y)) for all x,y in R,y!=0 and f(t)!=0, if t!=0 and f'(1)=3 then f(x) is

If int_(0)^(y)e^(-t^(2))dt+int_(0)^(x^(2))sin^(2)tdt=0 , then (dy)/(dx) at x=y=1 is