Question

Find derivatives of the given functions : x= ln (1+t^2) y= t- arctgt ?

Answer 1

`(dy)/(dx)=t/2`

Explanation:

When we have `f(t)=(x(t),y(t))`,

then `(dy)/(dx)=((dy)/(dt))/((dx)/(dt))`

Here `x(t)=ln(1+t^2)` and `(dx)/(dt)=(2t)/(1+t^2)`

and `y(t)=t-arctant` and therefore `(dy)/(dt)=1-1/(1+t^2)=t^2/(1+t^2)`

and hence `(dy)/(dx)=t^2/(2t)=t/2`

Answer 2

See details below

Explanation:

We know that if `f(x)=lnh(x)`, then `f´(x)=(h´(x))/(h(x))` and if `g(x)=arctanx`, then `g´(x)=1/(1+x^2)`

In our case, we have

`x(t)=ln(1+t^2)` here `h(x)=1+t^2`

`dx/dt=(2t)/(1+t^2)`

`y(t)=t-arctant`

`dy/dt=1-1/(1+t^2)`

Now if you look for `dy/dx=(1-1/(1+t^2))/(2t/(1+t^2))=t/2`