The First Derivative of a function which is defined parametrivally
as, x=x(t), y=y(t), is given by, dy/dx=(dy/dt)/(dx/dt); dx/dtne0...(ast)
Now, y=e^t rArr dy/dt=e^t, and, x=t^2+t rArr dx/dt=2t+1.
because, dx/dt=0 rArr t=-1/2, :., t ne-1/2 rArr dx/dt!=0.
:., by (ast), dy/dt=e^t/(2t+1), tne-1/2.
Therfore, (d^2y)/dx^2=d/dx{dy/dx},......."[Defn.],"
=d/dx{e^t/(2t+1)}
Observe that, here, we want to diff., w.r.t. x, a fun. of t, so, we
have to use the Chain Rule, and, accordingly, we have to first
diff. the fun. w.r.t. t and then multiply this derivative by dt/dx.
Symbolically, this is represented by,
(d^2y)/dx^2=d/dx{dy/dx}=d/dx{e^t/(2t+1)}
=d/dt{e^t/(2t+1)}*dt/dx
=[{(2t+1)d/dt(e^t)-e^td/dt(2t+1)}/(2t+1)^2]dt/dx
=[{(2t+1)e^t-e^t(2)}/(2t+1)^2]dt/dx
=((2t-1)e^t)/(2t+1)^2*dt/dx
Finally, noting that, dt/dx=1/{dx/dt},we conclude,
(d^2y)/dx^2=((2t-1)e^t)/(2t+1)^2*(1/(2t+1)), i.e.,
(d^2y)/dx^2=((2t-1)e^t)/(2t+1)^3, tne-1/2.
Enjoy Maths.!