Hi V k,

Here is the solution for your query

x=(1+t^2)^1/2

Now we need to differentiate this twice, that is double differentiate it to find acceleration.

x'=1/2(1/(1+t^2))^.5*2*t

=t/((1+t^2)^.5)

Differentiate using product rule:

x"= 1/((1+t^2)^3/2)

Dear VK,

x2=t2+1

taking its differentiation,

2x*(dx/dt)=2t

x*(dx/dt)=t

again taking differentiation for acceleration,

x*(d2x/dt2) + (dx/dt)2 = 1

x*(d2x/dt2) = 1 – (dx/dt)2

x*(d2x/dt2) = 1 – (t/x)2 (because x*(dx/dt)=t)

(d2x/dt2) = (1 – (t/x)2)/x

(d2x/dt2) = 1/x – t2/x3

put t2 = x2 -1

(d2x/dt2) = 1/x3

which is the required acceleration.

All the best

Hope this helps