What's the second derivative of x = t^2 + t, y = e^t?

1 Answer
Apr 20, 2017

(d^2y)/dx^2=((2t-1)e^t)/(2t+1)^3, tne-1/2.

Explanation:

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.!