Question

What is the derivative of `f(t) = ((lnt)^2-t, tsect )`?

Answer 1

The derivative of `f(t)=f'(t)=(2/t*lnt-1,sect(t*tant+1)).`

Explanation:

Let `f(t)=(x(t),y(t)),` so that `f'(t)=(x'(t),y'(t)).`

Now, `x(t)=(lnt)^2-t rArr x'(t)=d/dt{(lnt)^2-t}=d/dt(lnt)^2-d/dt(t)=2lnt*d/dt(lnt)-1=2/t*lnt-1.`

`y(t)=tsect rArr y'(t)=t*d/dt(sect)+sect*d/dt(t)=t*sect*tant+sect=sect(t*tant+1).`

Hence, the derivative of `f(t)=f'(t)=(2/t*lnt-1,sect(t*tant+1)).`