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

Question

2583 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-23T09:12:27

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

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 · 2018-04-23T09:13:43

See details below

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 $dy/dx=(1-1/(1+t^2))/(2t/(1+t^2))=t/2$