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.!
